LIBRARY 


UNIVERSITY  OF  CALIFORNIA. 


Class 


Carnegie  Uecbntcai  Scbools  TTeit  Boofcs 


MATHEMATICS 


FOR 


ENGINEERING   STUDENTS 


BY 


PROF.  S.  S.  KELLER 

CARNEGIE  TECHNICAL    SCHOOLS 


PLANE  AND  SOLID  GEOMETRY 


SECOND   EDITION,  REVISED 


NEW   YORK: 

D.  VAN    NOSTRAND    COMPANY 

23   MURRAY  AND   27   WARREN   STS. 


COPYRIGHT,   1907,   1908,  BY 
D.  VAN    NOSTRAND    COMPANY 


Stanbope  ipress 

F.    H.   GILSON     COMPANY 
BOSTON.      U.S.A.. 


PREFACE. 

IT  has  been  the  author's  endeavor  in  writing  this  book 
on  Geometry  to  put  the  student  at  ease  as  far  as  possible 
by  appealing  from  the  start  to  his  common  sense. 

Technicalities  have  been  avoided  wherever  possible 
and  the  student  encouraged  to  think  about  the  proposi- 
tions in  the  same,  simple,  common  sense  way  that  he 
would  consider  any  practical  question  arising  in  his 
daily  experience. 

This  process  is  applied  even  to  the  construction  of 
figures,  his  ingenuity  being  called  upon  for  suggestions 
as  to  auxiliary  lines,  etc.  In  this  way,  the  solution  of 
original  propositions  is  gradually  approached,  until 
confidence  is  acquired. 

In  this  work,  as  in  all  others,  the  author  has  adhered 
to  his  conviction  that  the  difficulty  in  mastering  mathe- 
matical truths  is  largely  due  to  the  student's  awe  of  the 
subject,  and  that  any  legitimate  means  of  securing 
confidence  and  of  removing  apprehension  should  be 
employed. 

S.  S.  K. 

Carnegie  Technical  Schools, 
Pittsburg,  Pa. 


227163 


PLANE   GEOMETRY 


DEFINITIONS. 

1.  GEOMETRY  is  a  study  of  position,  form,  and  dimension. 

2.  In  geometry  we  consider  points,  lines,  surfaces,  solids, 
and  angles. 

A  geometrical  point  has  only  position;  no  length  nor 
breadth.  Can  you  make  a  geometrical  point? 

Geometrical  lines  have  only  length;  no  width  nor  thick- 
ness. Can  you  draw  a  geometrical  line?  Give  a  defini- 
tion of  a  straight  line,  and  of  a  curved  line. 

Solids  have  length,  breadth,  and  height  or  thickness.  Can 
you  draw  a  solid? 

The  boundary  that  separates  a  solid  from  its  surroundings 
is  called  a  surface.  A  surface  has  extent,  but  no  thickness. 

3.  A  surface  is  called  a  plane  when  a  straight  line  joining 
any  two  points  in  it,  lies  wholly  within  the  surface.     Give 
an  illustration  of  a  surface  and  of  a  plane. 

4.  Plane  geometry  deals  only  with  figures  lying  in  the 
same  plane. 

5.  An  angle  is  the  amount  of  divergence  of  two  lines 
drawn  through  the  same  point.     The  point  is  called  the 
vertex  of  the  angle  and  the  two  lines,  its  sides. 

6.  A  segment  of  a  line  is  any  portion  of  it. 

When  a  line  is  divided  into  two  equal  segments,  it  is 
said  to  be  bisected.  When  cut  into  any  two  segments  it  is 
intersected. 

i 


«  ,l  -  \  Piane  Geometry. 


7.  A  segment  drawn  to  greater  length  is  said  to  be  pro- 
duced. 

8.  A  broken  line  is  made  up  of  a  number  of  straight  lines 
varying  in  direction. 

9.  When  two  straight   lines    cut  each  other  they  form 
four  angles  about   the  point  of  intersection.     The  angles 
situated  on  the  same  side  of  one  line  and  the  opposite  sides 
of  the  other  are  called  adjacent  angles  thus,  (see  Fig.  i  )  : 

ABC  and  CBD  are  adjacent  A. 
ABC  and  EBD  are  vertical  A. 

10.  When  one  straight  line  makes  equal  adjacent  angles 
with  another  it  is  said  to  be  perpendicular  to  it,  and  the 
equal  angles  are  called  right  angles,    xyw  and  wyv  (Fig.  2) 
are  right  angles. 


Fig.  i. 


Fig.  2. 


11.  A  straight  angle  is  an  angle  whose  sides  are  in  the 
same  straight  line  on  either  side  of  the  vertex,  thus  (Fig.  2 ) : 
xyv  is  a  straight  angle. 

Angles   are   read  with   the   letter  at   the   vertex  in  the 
middle, 
Thus:  angle  xyz  has  y  in  the  middle. 

angle  zyv  has  y  in  the  middle,  etc. 

12.  Angles  less  than  a  right  angle  are  called  acute;  -those 


Plane  Geometry.  3 

greater,  obtuse;  and  collectively,  they  are  known  as  oblique 
angles.  Hence  lines  meeting  at  any  angle  other  than  a 
right  angle  are  said  to  be  oblique  to  each  other. 

13.  If  two  angles  together  make  a  right  angle,  they  are 
said  to  be  complementary.     If  two  angles  together  make  a 
straight  angle,  they  are  said  to  be  supplementary,     wyz  and 
zyv  are  complementary,     xyz  and  zyv  are  supplementary 
(Fig.  2). 

AXIOMS. 

14.  The  demonstrations  of   geometry  are   based   upon 
certain  arbitrary  definitions  and  upon  certain  truths  that 
are  so  apparent  to  average  intelligence  that  they  require  no 
proof. 

Any  argument  that  arrives  at  a  conclusion  must  start 
from  some  truth  or  truths  accepted  by  the  parties  to  the 
argument. 

For  example,  no  amount  of  debate  on  the  phenomena 
of  sound  can  accomplish  any  result  if  one  disputant  starts 
from  the  definition  that  sound  is  vibration  in  the  air  and 
the  other  from  the  assertion  that  sound  is  the  effect  of  vibra- 
tion on  the  ear.  There  must  plainly  be  some  common 
ground  of  truth  from  which  both  must  reason.  So  in  geom- 
etry certain  fundamental  assumptions  must  be  agreed  upon 
before  reasoning  will  be  effective.  Hence  the  definitions 
just  suggested. 

Other  definitions  would  do  as  well,  if  they  were  universely 
accepted. 

Now  there  happen  to  be  certain  truths,  in  addition  to  these 
definitions,  that  are  so  evident  to  everyone  that  they  will  be 
readily  adopted.  These  truths  are  called  axioms  and  may 
be  stated  as  follows: 


4  Plane  Geometry. 

Axiom  i.  Things  which  are  equal  to  the  same  thing  are 
equal  to  each  other. 

Axiom  2.  If  equals  be  added  to  or  subtracted  from 
equals,  the  results  will  be  equal. 

Axiom  3.  If  equals  be  multiplied  or  divided  by  equals, 
the  results  are  equal. 

Axiom  4.  If  equals  be  added  to  or  subtracted  from 
unequals,  the  results  are  unequal  in  the  same  sense. 

Axiom  5.  If  unequals  be  substracted  from  equals,  the 
remainders  are  unequal  in  the  opposite  sense. 

Axiom  6.    The  whole  is  greater  than  any  of  its  parts. 
Axiom  7.    The  whole  is  equal  to  the  sum  of  all  its  parts. 

Axiom  8.  Through  two  points  only  one  straight  line  can 
be  drawn. 

Axiom  9.  The  shortest  distance  between  two  points  is 
the  straight  line  joining  them. 

Axiom  10.  Magnitudes  that  can  be  made  to  coincide  are 
equal,  etc. 

GEOMETRICAL   PROCESSES. 

15.  Geometrical  processes  are  of  two  kinds:  demon- 
stration and  construction. 

Demonstration  is  the  establishment  of  certain .  relations 
between  the  parts  of  figures  already  constructed. 

Construction  is  the  actual  method  of  building  those 
figures. 

A  statement  requiring  demonstration  is  called  a  theorem. 
A  statement  requiring  a  construction  is  called  a  problem. 
Proposition  includes  both. 


Plane  Geometry.  5 

A  truth  so  obvious  that  it  does  not  need  proof  is  called 
an  axiom  or  postulate,  the  latter  term  applying  to  purely 
geometrical  truths,  and  relating  more  to  methods. 

A  corollary  is  a  plain  inference  from  an  established 
theorem. 


POSTULATES. 

The  following  truths  may  be  assumed  without  proof. 
It  is  assumed  to  be  true : 

(a]  That  a  straight  line  can  be  drawn  from  any  one 

point  to  any  other  point. 

(b)  That  a  straight  line  can  be  produced  to  any  extent. 

ERPENDICULAR    AND    OBLIQUE    LINES. 

Proposition  I.     Theorem. 

16.   When  one  straight  line  crosses  another  straight  line 
the  vertical  angles  are  equal. 


Fig.  3. 

Let  line  OP  cross  AB  at  C. 

We  are  to  prove  Z  OCB  =  Z  ACP. 


6  Plane  Geometry. 

Analysis:  A  simple  inspection  of  Fig.  3  will  show  that 
both  ACP  and  OCB  combine  with  a  common  angle  AGO 
to  make  straight  angles,  which  are  always  equal. 
Proof:  Z  OCA  +  Z  OCB  =  2  rt.  A. 

(Being  sup.-adj.  A.) 
Z  OCA  +  ZACP  =  2  rt.  A. 
(Being  sup.-adj.  A.) 

.'.  Z  OCA  +  Z  OCB  =  Z  OCA  +  Z  ACP.  (Axiom.) 
Take  away  from  each  of  these  equals  the  common  Z  OCA. 
Then,  Z  OCB  -ZACP.  (Axiom.) 

In  like  manner  we  may  prove, 

ZOCA  =  ZBCP. 

Q.E.D 

Proposition  II.     Theorem. 

17.  When  the  sum  of  two  adjacent  angles  is  equal  to  two 
right  angles,  their  exterior  sides  form  one  and  the  same 
straight  line. 


Fig.  4. 

Let  the  adjacent  angles  Z  OCA  +  Z  OCB  =  2  rt.  A. 

We  are  to  prove  AC  and  CB  in  the  same  straight  line. 

Analysis:  It  is  evident  that  if  the  line  AC  be  extended 
sup. -adjacent  angles  will  be  formed  with  OC,  one  of  which 
will  be  AGO.  To  prove  that  CB  is  really  this  extension, 
it  is  only  necessary  to  prove  OCB  equal  to  the  angle  sup.- 
adjacent  to  AGO. 


Plane  Geometry 


Proof:  Suppose  CF  to  be  in  the  same  straight  line  with  AC. 
Then  Z  OCA  +  Z  OCF  =  2  rt.  Zs.)' 

(Being  sup. -adj.  A.) 
But  Z  OCA  +  Z  OCB  =  2  rt.  A. 

(By  hypothesis.) 
.-.  Z  OCA  +  Z  OCF  =  Z  OCA  +  Z  OCB. 

(Axiom. ) 

Take  away  from  each  of  these  equals  the  common  Z  OCA. 
Then  ZOCF  =  ZOCB. 

.'.  CB   and  CF  coincide,  and  cannot  form  two  distinct 
lines. 

/.  AC  and  CB  are  in  the  same  straight  line.       Q.E.D. 


Proposition  III.     Theorem. 

18.   A  perpendicular  measures  the  shortest  distance  from 
a  point  to  a  straight  line. 


E 
Fig-  5- 


Let  AB  be  the  given  straight  line,  C  the  given  point, 
and  CO  the  perpendicular. 

We  are  to  prove  CO  shorter  than  any  other  line  drawn 
from  C  to  AB,  as  CF. 


8  Plane  Geometry 

Analysis:  We  know  that  a  straight  line  is  the  shortest 
distance  between  two  points,  hence,  if  by  prolonging  each 
of  the  lines  CO  and  CF  an  equal  distance,  we  can  reduce 
the  comparison  to  one  between  a  straight  and  a  broken 
line,  our  proposition  is  easily  established. 

Produce  CO  to  E,  making  OE  =  OC.     Draw  EF. 

Proof:  On  AB  as  an  axis,  fold  over  OCF  until  it  comes 
into  the  plane  of  OEF. 

The  line  OC  will  take  the  direction  of  OE. 

(Since  Z  COF  =  EOF,  each  being  a  rt.  Z.) 

The  point  C  will  fall  upon  the  point  E. 

(Since     OC  =  OE  by  cons.) 

/.  Line  CF  =  line  FE, 
(having  their  extremities  in  the  same  points). 

.'.  CF  +  FE  =  2  CF, 
and  CO  +  OE  =  2  CO, 

But  CO  +  OE  <  CF  +  FE. 

(A  straight  line  is  the  shortest  distance  between  two 
points.) 

Substitute  2  CO  for  CO    +  OE, 

and  2  CF  for  CF  +  FE; 

then  we  have          2  CO  <  2  CF. 

/.  CO  <  CF. 

Q.E.D. 

Proposition  IV.     Theorem. 

19.  The  sum  of  two  lines  drawn  jrom  a  point  to  the 
extremities  of  a  straight  line  is  greater  than  the  sum  of  two 
other  lines  similarly  drawn,  but  included  by  them. 

Let  CA  and  CB  be  two  lines  drawn  from  the  point  C  to 
the  extremities  of  the  straight  line  AB.  Let  OA  and  OB 
be  two  lines  similarly  drawn,  but  included  by  CA  and  CB. 


Plane  Geometry 
We  are  to  prove  CA  +  CB  >  OA  +  OB. 


Fig.  6. 


Analysis:  In  comparison  of  magnitudes,  it  is  always 
desirable  to  have  some  known  connecting  bond.  Hence, 
if  one  of  the  included  lines  is  prolonged  until  it  meets  one  of 
the  including  lines,  we  have  straight  lines  and  broken  lines 
made  up  of  the  lines  involved  in  the  proposition,  which  can 
be  readily  compared. 

Proof:  Produce  AO  to  meet  the  line  CB  at  E. 

Then,  AC  +  CE  >  AO  +  OE, 

(A  straight  line  is  the  shortest  distance  between  two 
points),  and  BE  +  OE  >  BO. 

Add  these  inequalities  and  we  have, 

AC  +  CE  +  BE  +  OE  >  OA  +  OE  +  OB. 

Substitute  for  CE  +  BE  its  equal  CB,  and  take  away 
OE  from  both  sides  of  the  inequality,  we  have, 
CA  +  CB  >  OA  +  OB. 


Proposition  V. 

20.    Two  oblique  lines  drawn  from  the  same  point  on  a 
perpendicular  to  a  straight  line,  cutting  off  equal  distances 


10 


Plane  Geometry 


\rorn  its  foot,  are  equal;  and  of  two  oblique  lines,  drawn 
from  the  same  point  on  a  perpendicular,  cutting  off  unequal 
distances,  the  more  remote  is  the  greater. 

Statement:   To    prove   that    MN  =  MP,    if   NO  =  OP, 
and  that  MR  >  MN  if  OR>NO. 


Fig.  7- 


Analysis:  MON  and  MOP  being  right  angles  enable  us 
to  use  MO  as  an  axis  upon  which  to  revolve  MON  around 
upon  MOP.  Also  NO  being  equal  to  OP,  gives  a  ready 
method  of  determining  the  position  of  the  point  N  relative 
to  point  P. 

Proof:  (Part  i.)  Fold  over  MOP  on  MO  as  an  axis. 
Since  MOP  and  MON  are  both  right  angles  (because  MO 
is  a  J_),  OP  will  fall  on  NO,  and  since  OP  -  NO,  the 
point  P  will  fall  on  the  point  N;  hence  MP  will  exactly  coin- 
cide with  MN,  since  MP  and  MN  have  the  point  M  in 
common,  and  the  point  P  exactly  coinciding  with  N. 
/.  MP  =  MN. 

(Part  2.)  To  prove  MR  >  MN,  if  OR  >  NO.  Since 
MP  =  MN,  it  will  be  sufficient  to  prove  MR  >  MP. 

Analysis.  Prop.  IV  suggests  the  method  of  extend- 
ing the  line  MO  below  NR  and  joining  the  extremity  of 
this  produced  line  with  P  and  R,  thus  reproducing  the  con- 


Plane  Geometry  II 

ditions  of  Prop.  IV.  It  is  plainly  desirable  to  make  the 
figure  as  simple  as  possible  by  producing  MO  a  distance 
OT  equal  to  MO. 

Proof:   Joining  T  with  P  and  R,  we  have 

PT  =  PM  )  (by  part  i,  since  OR  is -L  to  MT 
and  RT  =  RM  j    and  MO  =  OT) 

RM  +  RT  >  PM  +  PT  (Prop.  IV.) 

or          2  RM  >  2  PM     (since  RM  =  RT  and  PM  =  PT.) 

.-.     RM  >  PM. 
That  is,  RM  >  MN. 

21.  Cor.  I.  Since  in  Part  i,  the  two  two  triangles  MOP 
and  MON  exactly  coincided  when  MOP  was  folded  over 
upon   MON,    the    Z  OMP  =  Z  OMN;  that   is,    oblique 
lines  drawn  from  the  same  point  on  a  perpendicular  to  a 
straight  line,  cutting  off  equal  distances  from  its  foot,  make 
equal  angles  with  the  perpendicular. 

Proposition  VI.     Theorem. 

22.  //  at  the  middle  point  of  a  straight  line  a  perpendicu- 
lar be  erected : 


Pig.  8 

(a)  any  point  in  the  perpendicular  is  at  equal  distances 
from  the  extremities  of  the  straight  line. 


12 


Plane  Geometry 


*    (b)  any  point  without  the  perpendicular  is  at  unequal 
distances  from  the  extremities  oj  the  straight  line. 

Let  PR  be  a  perpendicular  erected  at  the  middle  of  the 
straight  line  AB,  O  any  point  in  PR  and  C  any  point  with- 
out PR. 

(a)  Draw  OA  and  OB.     We  are  to  prove  OA  =  OB. 
Since  PA  =  PB,  OA  =  OB   (two  oblique   lines  drawn 

from  the  same  point  in  a  _L,  cutting  off  equal  distances  from 
the  foot  of  the  _L  are  equal.) 

(b)  Draw  CA  and  CB;  C  being  a  point  outside  PR.   We 
are  to  prove  CA  and  CB  unequal. 

One  of  these  lines,  as  CA,  will  intersect  the  _L  at  D. 
From  the  point  of  intersection  draw  DB.  DB  =  DA. 
(Two  oblique  lines  drawn  from  the  same  point  in  a  J_,  cut- 
ting off  equal  distances  from  the  foot  of  the  _L,  are  equal.) 
CB  <  CD  +  DB  (a  straight  line  is  the  shortest  distance 
between  two  points).  Substitute  for  DB  its  equal  DA, 
then  CB  <  CD  +  DA. 

But  CD  +  DA  =  CA. 

.;.  CB  <  CA.  Q.E.D. 

Proposition   VII. 

23.  But  one  perpendicular  can  be  drawn  to  a  given  line 
through  a  given  point. 


Fig  9. 


Statement:  PN  is  the  only  perpendicular  through  P  to  AB. 


Plane  Geometry.  13 

Analysis:  It  is  known  that  but  one  straight  line  can  be 
drawn,  joining  two  given  points;  hence  by  producing  PN 
below  AB  and  joining  the  extremity  with  the  foot  of  any 
other  line,  presumed  to  be  also  a  perpendicular,  it  will  be 
possible  to  show  that  PN  produced  is  the  only  straight  line, 
between  the  point  P  and  this  other  extremity. 

Proof:  Draw  any  other  line  PM,  to  show  that  it  is  not 
a  perpendicular.  Produce  PN  to  O,  making  NO  =  NP, 
and  join  O  with  M.  Then  since  only  one  straight  line  can 
be  drawn  between  P  and  O,  OMP  cannot  be  a  straight  line, 
and  hence  Z  PMO  cannot  be  a  straight  angle. 

But  by  Cor.  I,  Prop.  V,  Z  PMN  =  Z  OMN  (for  AB  is 
_L  to  OP). 

.-.  ZPMN=  iZPMO. 

.'.  If  ZPMO  is  not  a  straight  angle,  Z  PMN  is  not  a 
right  angle.  Hence  PM  is  not  J_  to  AB. 

Since  PM  is  any  line  other  than  PN,  PN  is  the  only  J_. 

Proposition  VIII.     Theorem. 

24.  Two  straight  lines  in  the  same  plane  perpendicular 
to  the  same  straight  line  are  parallel. 

M 
A B 


K 

Fig.  10. 

Statement:  Let  AB  and  CD  be  both  _L  tc  MN,  then  are 
they  parallel. 


14  Plane  Geometry. 

Analysis:  Parallel  lines  can  never  meet  no  matter  how 
far  produced.  If  AB  and  CD  do  not  meet,  then  they  must 
be  parallel. 

Proof:  If  AB  and  CD  should  meet  when  sufficiently 
prolonged,  they  would  then  both  extend  from  that  common 
point  of  meeting,  say  E,  to  the  line  MN,  thus  (Fig.  1 1 ) : 


M 

A 


Fig.  ix. 

and  since  they  are  both  _L  to  MN  by  hypothesis,  we  would 
have  two  _L's  drawn  from  the  same  point  to  the  same  straight 
line,  which  we  know  to  be  impossible,  .'.  they  cannot  meet, 
and  hence  are  parallel. 

Proposition  IX. 

25.   //  a  straight  line  is  perpendicular  to  one  of  two  par- 
allel lines,  it  is  perpendicular  to  the  other  also. 


Fig.  12. 

Statement:  Let  MN  be  _L  to  AB,  which  is  parallel  to  CD, 
to  prove  MN  _L  to  CD 


Plane  Geometry.  15 

Analysis:  A  line  drawn  through  O-L  to  MN  will  be  ||  to 
AB  by  last  proposition.  If  this  line  can  be  identified  with 
CD,  our  conclusion  follows. 

Proof:  Through  O  where  MN  meets  CD,  draw  a  line 
EF  J_  to  MN,  and  entirely  independent  of  CD ;  then  by  last 
proposition  AB  and  EF  are  ||.  But  CD  is  already  drawn 
through  O  ||  to  AB  by  our  hypothesis,  and  since  through 
one  point  it  is  impossible  to  draw  more  than  one  line  ||  to 
another  given  line,  EF  and  CD  must  be  the  same  line.  But 
EF  was  drawn  J_  to  MN,  .'.  CD  is  _L  to  MN. 

26.  Definition:  A  line  cutting  two  or  more  straight  lines 
is  called  a  transversal. 


HD  is  a  transversal.  Z  ABF  and  Z  BFG  ;  also 
ZCBF  and  BFE  are  called  alternate-interior  angles. 

HBC  and  BFG;  also  HBA  and  BFE  are  called  exterior- 
interior  angles.  HBC  and  EFD;  also  ABH  and  DFG  are 

called  alternate-exterior  angles. 


Proposition  X. 

27.   //  a  transversal  cuts  two  parallel  lines,  the  alternate- 
interior  angles  thus  formed  are  equal;  the  interior -exterior 


16  Plane  Geometry. 

angles  are  equal  and  the  interior  angles  on  the  same  side  of 
the  transversal  are  supplementary. 

Statement:    If  a  transversal  MN  cuts  the  parallels  AD 
and  EH,  the  angle  DBG  =  angle  BGE;  the  angle  DBM  = 
angle  HGB  (or  ABM  =  BGE)  and  the  angle  DBG  +  angle 
BGH  =  two  right  angles,  etc. 


H- 


Fig.  14. 

Analysis:  To  prove  two  angles  equal  it  is  necessary  first 
to  prove  that  they  will  exactly  coincide  when  applied  to 
one  another;  and  to  apply  them  effectively  it  is  best  to 
inclose  them  in  a  triangle. 

Proof:  Draw  the  auxiliary  line  CF  through  the  middle 
point  O  of  BG  and  _L  to  AD,  hence  J_  to  EH.  (Why?) 
What  reason  is  there  for  drawing  CF  through  the  middle 
ofBG? 

In  order  to  compare  Z  CBO  (DBG)  with  Z  OGF  (BGE), 
to  which  we  are  to  prove  it  equal,  swing  the  triangle  OFG 
about  its  vertex  O  as  a  pivot,  until  OG  falls  directly  upon 
OB.  Then,  since  O  was  the  mid-point  of  BG,  OG  must 
equal  OB  and  the  point  G  will  fall  exactly  upon  B,  because 
O  is  fixed. 


Plane  Geometry.  17 

Since  ZFOG=ZBOC  (why?),  OF  will  fall  upon  OC 
(the  other  sides  of  these  angles,  OG  and  OB  already  co- 
inciding). 

Since  FG  and  BC,  being  parts  of  EH  and  AD  respec- 
tively are  both  J_  to  CF  (which  is  simply  folded  upon  itself); 
since  G  is  now  at  B  and  but  one  perpendicular  can  be 
drawn  from  a  given  point  to  a  given  straight  line,  FG 
must  fall  upon  BC,  and  since  two  straight  lines  can  meet 
at  but  one  point,  .'.  F  must  coincide  with  C,  .'.  the  two  tri- 
angles BOC  and  OFG  coincide  throughout  and  Z  CBO  = 
Z  OGF,  or  Z  DBG  =  Z  BGE. 

Since     Z  GBD  +  Z  GBA  =  2  rt.  /4, 
and  Z  BGE  +  BGH  =  2  rt.  Zs, 

/.  GBD  +  GBA  =  BGE  +  BGH.   (Axiom.) 
But,  GBD  =  BGE. 

Subtracting;  GBA  -  BGH.  (Axiom.) 

Again,  to  prove 

Z  DBM  =  Z  HGB ;     Z  GBA  =  DBM.   (Why  ?) 

and  r     ZGBA=ZBGH.  (Why?) 

/.  DBM  -  Z  BGH.  (Axiom.) 

Again,  to  prove  ZDBG  +  Z  BGH  =  2  right  angles. 

ABG  +  GBD  -  2  right  angles.          (Why?) 

But  ABG  =  BGH. 

/.  BGH  +  DBG  =  2   right  angles. 

(Axiom.) 

Proposition  XI.      (Conversely.) 

29.  When  a  transversal  cuts  two  lines  so  that  the  alter- 
nate-interior angles  are  equal,  the  lines  are  parallel. 

Statement:  Let  xy  cut  AB  and  FH  so  that  Z  ACG  = 
Z  CGH,  then  AB  is  ||  to  FH. 


1 8  Plane  Geometry. 

Analysis :  If  a  line  drawn  through  C,  the  point  of  in- 
tersection of  AB  with  xy,  parallel  to  FH  and  AB,  can 
be  identified  with  this  line,  our  proposition  is  established. 
Prop.  X  will  aid  the  identification. 


_B 


-H 


y 

Fig.  15- 

Proof:  Draw  the  line  DE  through  C  ||  to  FH. 

Then  Z  DCG  =  Z  CGH.  (By  Prop.  X.) 

But  Z  ACG  -  Z  CGH.  (By  hypothesis) 

.'.  ZDCG  -  ZACG.  (Axiom.) 

Since  CG  is  a  side  common  to  both  angles  (DCG  and 
ACG),  the  other  sides  (DC  and  AC)  must  coincide.  /.  whole 
line  AB  coincides  with  whole  line  DE.  But  DE  was 
drawn  ||  to  FH.  .'.  AB  is  ||  to  FH. 

Triangles. 

30.  Definitions:  A  triangle  is  a  -portion  of  a  plane  in- 
closed by  three  straight  lines. 

31.  The  bounding  lines  are  called  sides,  and  their  sum, 
the  perimeter  of  the  triangle.     The  points  of  intersection 
of  the  lines  are  called  vertices.     The  adjacent  parts  of  a 
triangle  (consisting  of  angles  and  sides)  are  those  between 
which  no  other  part  intervenes. 


Plane  Geometry.  19 

32.  The  angle  formed  outside  a  triangle  by  one  of  its 
sides  and  the  prolongation  of  the  adjacent  side  is  called  an 
exterior  angle:  thus,  CAB,  ECD,  and  FEA  are  all  exterior 
angles.  The  two  angles  within  the  triangle  not  adjacent 
to  an  exterior  angle,  are  called  its  opposite  interior  angles. 


Fig.  16. 

33.  A  triangle  is  called  scalene  when  no  two  of  its  sides 
are  equal;  isoceles  when  two  of  its  sides  are  equal;  equila- 
teral when  all  three  sides  are  equal.     A  right  triangle  has 
one  right  angle.     Define  an  equiangular  triangle. 

34.  In  a  right  triangle,  the  side  opposite  the  right  angle 
is  called  the  hypotenuse,  and  the  other  two  sides,  the  legs. 

35.  Any  side  of  a  triangle  may  be  called  its  base.     It  is 
customary  to  call  the  equal  sides  of  an  isosceles  triangle, 
the  legs  and  the  'other  side,  the  base. 

36.  The  angle  opposite  the  side  upon  which  the  triangle 
rests  (the  base)  is  called  the  vertical  angle. 

37.  The  altitude  of  a  triangle  is  the  perpendicular  dis- 
tance from  its  vertex  to  its  base.     The  lines  drawn  from 
the  vertices  of  a  triangle  to  the  middle  of  the  opposite  sides 
are  called  medians. 

38.  The  parts  similarly  situated  in  different  figures  are 
called  homologous  parts. 


20  Plane  Geometry. 

39.  Any  two  or  more  figures  are  said  to  be  equal  when 
they  can  be  made  to  exactly  coincide  when  applied  to 
each  other. 


Proposition  XII. 

40.  The  sum  of  the  three  angles  oj  any  plane  triangle  is 
equal  to  two  right  angles. 

Statement:  In  the  triangle  ABC,  Z  ABC  +  Z  BAG  + 
ZACB  =  2  rt.  Zs. 


F.g.  17. 

Analysis:  All  the  angles  formed  on  one  side  of  a  straight 
line  about  one  point  together  make  two  right  angles.  If 
we  can  prove  that  the  angles  of  any  triangle  equals  three 
angles  formed  about  one  of  its  vertices,  on  the  same  side 
of  one  of  its  sides  (for  instance  about  the  vertex  C  above 
BC  or  BC  produced),  the  proposition  is  proved. 

To  do  this  we  take  advantage  of  the  proposition  estab- 
lishing the  equality  of  angles  formed  by  parallel  lines,  by 
drawing  the  auxiliary  line  CD  ||  to  AB  and  by  producing 
BC,  say  to  E. 

Proof:  Since  BA  and  CD  are  ||  and  are  cut  by  the  trans- 
versal, a  portion  of  which  is  the  side  AC,  Z  BAG  =  the 


Plane  Geometry.  21 

alternate  interior  angle  ACD;  also  AB  and  CD  are  cut  by 
the    transversal   BE  and    Z  ABC  =  its   opposite   exterior 
Z  DCE. 
Then 

ZACB  +  ZACD  +  ZDCE=ZACB  +  ZBAC+ZABC. 
But 

Z  ACB  +  ZACD  +  ZDCE=2  rt.  A.        (Why?) 
/.ZACB  +  ZBAC+ZABC-2  rt.  A.        (Why?) 

41.  Cor.  I.     If  two  angles  of  a  triangle  (or  their  sum) 
are  known,  the  third  angle  may  be  found  by  subtracting 
their  sum  from  two  right  angles. 

42.  Cor.    II.     If  two  triangles  have  two  angles  of  one 
(or  their  sum)  equal  to  two  angles  of  the  other  (or  their 
sum)  the  third  angles  are  equal. 

43.  Cor.  III.     If    two    right    triangles    have    one    acute 
angle  in  each  equal,  the  other  acute  angles  are  equal. 

44.  Cor.  IV.     There  cannot  be  two  right  angles  in  one 
triangle.  (Why?) 

45.  Cor.   V.     In  a  right  triangle  the  two  acute  angles 
are  complementary. 

46.  Cor.  VI.     An  exterior  angle  of  a  triangle  is  equal  to 
the  sum  of  the  two  opposite  interior  angles. 


Proposition  XIII. 

47.    The  difference  of  two  sides  of 
a  triangle  is  less  than  the  third  side.  pig.  18. 

Statement  AG  -  AB  <  BC. 

Proof:    Since   a  straight    line  is  the  shortest   distance 

between   two   points   AC  <  AB  +  BC.  Substracting   AB 
from  each  side  leaves  AC  —  AB  <  BC.  (Axiom.) 


22  Plane  Geometry. 


Proposition  XIV. 

48.  Two  triangles  are  equal  if  two  angles  and  their  com- 
mon side  in  one,  equals  two  angles  and  their  common  side 
in  the  other. 

Statement:  A  ABC  =  A  DEF,  if  Z  BAG,  Z  BCA 
and  side  AC  in  A  ABC  equal  Z  EDF,  Z  EFD  and  side 
DF  in  A  DEF. 


Fig.  19. 


Analysis:  Since  the  only  test  of  equality  is  coincidence, 
we  must  apply  one  triangle  to  the  other  and  compare.  We 
know  that  AC  can  be  made  to  coincide  exactly  with  DF, 
A  being  at  D. 

Proof:  Since,  AC  =  DF,  C  will  fall  exactly  on  F.  Since 
Z  BAG  =  Z  EDF,  AB  will  take  the  direction  of  DE, 
from  definition  of  angle.  For  like  reason  BC  will  take  the 
direction  of  EF,  and  hence  they  must  meet  at  the  same 
point  E. 

.-.  B  falls  on  E,  and  the  two  triangles  exactly  coincide, 
hence  they  are  equal. 

49.  Cor  I.  Two  triangles  are  equal  if  a  side  and  any 
two  angles  of  one  are  equal  to  a  side  and  any  two  angles  of 
other.  (Why?) 

What  parts  must  be  equal  in  a  right  triangle,  that  the 
triangles  may  be  equal? 


Plane  Geometry.  23 

Proposition  XV. 

50.  Two  triangles  are  equal,  if  two  sides  and  the  angle 
they  jorm  in  one,  are  equal  to  two  sides  and  the  angle  they 
form  in  the  other. 

Statement:  A  ABC  =  A  DEF,  if 

AB,  AC  and  Z  BAG  in  A  ABC  =  DE,  DF  and  Z  EDF 
in  A  DEF. 


Fig.  20. 

Analysis:  The  final  test  of  equality  is  coincidence  and 
one  triangle  must  be  applied  to  the  other,  making  one  of 
the  equal  parts  in  each  coincide  at  the  start. 

Proof:  Place  DEF  upon  ABC,  making  DF  fall  upon  AC. 
Since  DF  =  AC,  if  D  is  placed  upon  A,  F  will  fall  upon  C. 
And  since  Z  EDF  -  Z  BAG,  DE  will  fall  upon  AB, 
also  DE  being  equal  to  AB,  E  will  fall  upon  B.  Since  the 
lines  EF  and  BC  have  two  points  each  in  common, 
namely  B  arid  E,  and  F  and  C,  EF  will  coincide  with  BC. 

.'.  ABC  and  DEF  coincide  throughout,  hence  are  equal. 

Proposition  XVI. 

51.  Two  triangles  which  have  three  sides  of  the  one 
equal  to  three  sides  of  the  other  are  equal. 

Statement:  A  ABC  =  Axyz  if  AB  =  xy,  BC  =  yz  and 
AC  =  xz. 


24  Plane  Geometry. 

Analysis:  Equality  is  always  proved  finally  by  coinci- 
dence, but  as  no  angles  are  given  in  this  case,  we  cannot 
determine  the  relative  directions  of  the  sides.  WTe  must 
then  resort  to  some  plan  that  will  enable  us  to  find  at  least 
one  angle. 


Instead  of  placing  one  triangle  directly  over  the  other,  we 
apply  a  side  of  one  to  its  homologous  side  in  the  other,  mak- 
ing them  exactly  coincide  and  then  compare  the  other  parts. 

Proof:  Place  the  side  AC  on  the  side  xy,  so  that  A  falls 
on  x  and  C  on  z,  and  the  vertex  B  falls  on  the  opposite  side 
of  xy  from  y,  thus  (Fig.  22.): 


2(0) 


rig.  22. 


Join  the  vertices  y  and  B. 

Now  since  xy  =  AB  by  hypothesis,  xy  =  ocE  (since  x 
and  A  are  same  point  now)  in  the  triangle  "Bxy,  and 
Z*;yB  =  Z.  xEy.  (Why?) 


Plane  Geometry. 


25 


Likewise  yzB  is  an  isosceles  triangle  and  Z  B^z  =  /_  yEz. 

But  ZxyB=ZxBy.        (By  above.) 

Add  ;  /_  Eyz  +  Z  xyE  =  ZyEz+  Z  xfty, 

or  Z  xyz  =  Z  xEz  =  Z  ABC. 

Then  in  the  two  triangles  xyz  and  ABC  we  have  two  sides 
(AB  and  BC)  in  one  =  to  two  sides  (xy  and  yz)  in  the 
other,  and  the  angle  ABC  between  AB  and  BC  in  ABC  = 
Z  xyz  between  xy  and  yz  in  xyz,  hence  the  two  triangles 
are  equal. 

52.   Cor.  I.   Two  right  triangles  are  equal  if  their  legs 
are  equal. 


Proposition  XVII. 

53.  In  an  isosceles  triangle,  the  angles  opposite  the  equal 
sides  are  equal ;  and  conversely,  if  in  any  triangle  two  of  the 
angles  are  equal  the  triangle  is  isosceles. 

Statement  of  first  part: 

In  the  isosceles  triangle  MNO,  having  MN  =  NO,  the 
Z  NMO  -  Z  NOM. 


Fig.  23. 


Analysis:  It  is  plainly  necessary  to  divide  the  triangle 
into  two  parts  to  be  proved  equal;  and  that  can  be  done 
only  by  a  line  from  vertex  N. 


26 


Plane  Geometry. 


Proof:  Draw  NP  bisecting  /_  MNO,  then  we  have  two 
triangles  having  the  sides  MN  and  NO  equal,  the  side  NP, 
common  to  both,  and  the  angle  MNP  =  angle  ONP  (halves 
of  same  angle),  hence  the  A's  are  equal.  (Why?) 

/.  Z  NMP  =  homologous  Z  NOP. 

Statement  of  second  part: 
Let  Z  A  -  Z  C  in  triangle  ABC. 

To  prove  AB  =  BC. 

Same  analysis  as  first  part. 


Fig.  24- 

Proof:  Draw  BD  _L  to  base  AC.  Then  in  the  two  right 
triangles  ABD  and  DBC,  BD  is  common  to  both  and  angle 
BAD  =  angle  BCD  (by  hypothesis),  hence  the  triangles 
are  equal.  (Why?) 

Hence,  AB  =  BC  and  ABC  is  isosceles. 

54.  Cor.  I.    Since  AD  also  equals  DC  and  Z.ABD  - 
Z  DBC  (because  triangle  ABD  and  BDC  are  equal),  the 
perpendicular  from  the  vertex  of  an  isosceles  triangle   to  the 
base  bisects  the  base  and  also  the  vertical  angle. 

55.  Cor.  II.   An  equiangular  triangle  is  also  equilateral. 


Plane  Geometry. 
Proposition  XVIII. 


27 


56.  Two  right  triangles  are  equal  if  a  leg  and  the  hypo- 
tenuse of  one  is  equal  to  a  leg  and  the  hypotenuse  of  the 
other. 


Fig.  25. 

Statement:  If  ABC  and  DEF  have  AB  =  DF  and  AC  = 
DE,  then  ABC  =  DEF. 

Analysis:  This  is  plainly  a  case  for  comparison  of  figures, 
by  some  form  of  superposition. 

Proof:  Apply  DEF  to  ABC,  making  the  equal  sides  DE 
and  AC  coincide  exactly,  F  and  B,  being  on  opposite  sides 
of  this  common  line.  Then  BC  and  CF  (EF)  will  form 
one  straight  line.  (Why?) 

Then  the  triangle  BAF  is  isosceles  because  AB  =  DF 
(now  AF)  by  hypothesis;  hence,  Z  ABF  =  Z  AFB,  that 
is,  the  two  right  triangles  have  two  sides  and  an  acute  angle 
equal  in  each,  .*.  the  triangles  are  equal. 


Proposition  XIX. 

57.  //  two  sides  of  a  triangle  are  unequal,  the  angle 
opposite  the  greater  side  is  greater  than  the  angle  opposite 
the  smaller  side. 

Statement:  If  BC  >  AB  in  the  triangle  ABC,  Z  BAG  > 
Z  BCA. 


28  Plane  Geometry. 

Analysis:  A  comparison  of  the  isosceles  triangle  formed 
by  laying  off  a  distance  on  the  greater  side  BC,  equal  to 
AB,  and  joining  this  point  with  A,  will  show  the  relation 
between  the  angles,  if  we  observe  that  one  of  the  angles  of 
this  isosceles  triangle  is  the  exterior  angle  of  the  remaining 
triangle. 


Fig.  26. 

Proof:  Lay  off  BD  =  AB  and  draw  the  auxiliary  line 
AD.  Then  ABD  is  an  isosceles  triangle,  and  hence, 
Z  BAD  =  Z  BDA.  But  Z  BDA,  being  the  exterior 
angle  of  the  triangle  ADC,  is  equal  to  the  sum  of  Z  DAG 
and  Z  DCA,  hence  is  greater  than  either  of  them,  that  is 
ZBDA  >  ZDCA,  but  Z  BDA  =  Z  BAD,  which  is 
plainly  less  than  BAG,  since  it  is  only  a  part  of  Z  BAG. 
That  is,  since  Z  BDA  =  Z  BAD,  Z  BDA  <  Z  BAG; 
but  Z  BDA  >  Z  DCA  or  Z  DCA  <  Z  BDA,  therefore, 
Z  DCA  being  less  than  Z  BDA,  which  is  itself  less  than 
Z  BAG,  Z  DCA  (or  Z  BCA)  must  be  less  than  Z  BAG. 

Proposition  XX. 

68.  Reciprocally:  If  two  angles  of  triangle  are  unequal, 
the  greater  side  is  opposite  the  greater  angle. 

Statement :  If  in  Z  ABC,  Z  ABC  is  greater  than  ACB, 
the  side  AC  >  side  AB. 


Plane  Geometry. 


29 


Analysis  and  Proof :  AB  is  either  -equal  to,  greater  than, 
or  less  than  AC;  there  is  no  other  alternative. 

If  AB  =  AC,  the  triangle  is  isosceles  and  angle  ABC  = 
angle  ACB,  which  is  not  true. 


Fig.  27. 

If  AB  >  AC,  by  the  last  proposition  Z  ACB,  opposite 
AB  is  greater  than  Z  ABC,  opposite  AC,  but  the  proposi- 
tion says  Z  ABC  >  Z  ACB,  which  leaves  us  but  the  one 
alternative,  that  AB  <  AC. 

Proposition  XXI. 

59.  //  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other,  but  the  angles  formed  by  them  unequal, 
then  the  third  side  in  the  triangle  having  the  greater  angle, 
is  the  greater. 


Fig.  28. 


Statement:  If  in   triangles  ABC   and  DEF,  AB  =  DE, 
BC  -  EF,  and  Z  DEF  >  Z  ABC,  then  DF  >  AC. 


30  Plane  Geometry. 

Analysis :  The  two  triangles  must  be  compared  directly 
since  it  is  a  case  of  inequality,  and  hence  one  must  be 
applied  to  the  other,  and  auxiliary  lines  employed  to  get 
the  relation. 

Proof :  Apply  ABC  to  DEF  making  AB  coincide  with 
its  equal  DE;  then  Z  ABC  being  less  than  Z  DEF,  BC 
will  fall  inside  of  DEF  as  in  the  figure. 

Draw  the  auxiliary  line  EG  bisecting  Z  CEF,  meeting 
DF  in  G,  join  C  and  G. 

NOTE.  —  The  wisdom  of  this  construction  is  apparent, 
if  it  be  observed  that  DC  and  DF  are  to  be  compared. 
Hence  if  a  point  on  DF  can  be  found  (as  G)  such  that  if 
the  segment,  GF,  of  DF  can  be  turned  about  G  so  that 
F  falls  on  C,  then  we  have  a  straight  line  DC  compared 
with  a  broken  line,  which  is  equal  to  DF.  This  point  G 
can  only  be  found  by  bisecting  Z  CEF. 

In  A's  EGF  and  EGC,  EF  -  EC  (by  hypothesis),  EG 
is  common  and  Z  GEF  between  EF  and  EG  in  A  EGF, 
equals  <  GEC  between  EG  and  EC  in  A  ECG,  hence 
A  EGF  -  A  ECG,  and  CG  =  GF  (the  other  sides  in 
the  A's  EGF  and  EGC). 

In  DGC,  DG  +  GC>  DC.  (Why?) 

But      GC  =  GF  .-.  DG  +  GF  >  DC, 
that  is,     DF  >  DC,  or  DF  >  AC. 

Proposition  XXII. 

60.  State  and  prove  the  converse  of  Prop.  XXI. 

Proposition  XXIII. 

61.  The  perpendicular  bisector  of  a  line  is  the  locus  of 
points  equidistant  from  the  extremities  of  the  line.    • 


Plane  Geometry.  31 

Statement:  If  xy  is  the  perpendicular  bisector  of  line  AC, 
it  is  the  locus  of  points  equidistant  from  A  and  C. 


y 

Fig.  29- 

Analysis:  To  show  that  xy  is  the  specified  locus,  it  is 
necessary  to  prove  that  every  point  on  it  is  equidistant 
from  A  and  C,  and  also  that  any  point  outside  of  xy  is 
unequally  distant  from  A  and  C. 

Proof :  Let  B  be  any  point  on  the  _L  bisector  xy,  then  in 
the  two  right  triangles  ABy  and  yBC,  By  is  common  and 
Ay  =  yC  because  y  is  the  middle  point  of  AC.  Hence 
triangle  ABy  =  triangle  yBC  (Why?).  Hence  AB  -  BC. 
That  is,  B  (any  point  on  xy)  is  equally  distant  from  A 
and  C. 

2d.   Let  R  be  any  point  outside  of  xy,  then  drawing  AR, 
RC,  and  OC,  O  being  the  point  where  AR  cuts  xy,  AO  = 
OC   by  first   part,    and   OR  +  OC  >  RC,   that  is,   since 
OC  -  AO,  OR  +  AO  >  RC  or  AR  >  RC. 


Proposition  XXIV. 

62.    The   lorn     of  points   within  an  angle  and  equally 
distant  from  the  sides  is  the  bisector  of  the  angle. 


32  Plane  Geometry. 

Statement:  Let  A  be  any  angle  ;  then  is  AM,  the  bisector, 
the  locus  of  points  within  the  angle  A  equally  distant 
from  its  sides. 


Fig.  30. 

Analysis:  To  prove  that  AM  is  the  required  locus,  it 
is  necessary  to  show  that  every  point  on  AM  is  equally 
distant  from  Ax  and  Ay  and  that  any  point  not  on  AM 
is  unequally  distant;  or  that  every  point  on  AM  is  equally 
distant  from  Ax  and  Ay  and  that  every  point  equally  dis- 
tant from  Ax  and  Ay  is  on  AM. 

In  the  first  process,  then,  we  must  take  any  point  on 
AM  and  prove  that  the  right  triangles  formed  by  perpen- 
diculars drawn  from  it  to  the  sides  are  equal,  and  taking 
a  point  without  to  show  that  the  perpendiculars  drawn 
from  it  to  the  sides  are  unequal. 

Proof:  Let  C  be  any  point  on  AM.  Draw  the  perpen- 
diculars CB  and  CD.  Then  in  the  two  right  triangles 
ABC  and  ACD,  AC  is  common  and  Z  BAG  =  Z  CAD 
being  halves  of  same  angle,  therefore,  ZACB  =  ZACD 
(Why?),  and  we  have  two  angles  and  the  included  side, 
.'.  triangle  ABC  =  triangle  ACD,  hence  side  BC  =  side  CD. 

2d.  Take,  G,  any  point  outside  of  AM,  and  draw  the 
perpendiculars  GE  and  GK  to  the  sides,  then  GK  <  GE. 


Plane  Geometry.  33 

Draw  FR  _L  Ay  from  F  —  the  point  where  GE  cuts  AM — 
also  RG.  (Why?)  /.  FE  +  FG  >  GR  (since  FE  = 
FR)  or,  since  FE  +  FG  =  EG,  EG  >  GR.  But  GR  > 
GK.  (The  perpendicular  is  shortest  distance  from  a 
point  to  a  line.)  .'.  EG  >  GK. 

Exercise:  As  an  exercise  prove  that  every  point  equally 
distant  from  Ax  and  Ay  is  on  the  bisector  AM. 


QUADRILATERAL. 

Definitions. 

A  quadrilateral,  as  the  name  signifies,  is  a  four  sided 
figure. 

Any  side  of  a  quadrilateral  may  be  called  its  base. 

The  diagonal  of  any  quadrilateral  is  the  straight  line 
joining  any  two  vertices  not  adjacent. 
"  The  vertices  are  the  points  where  the  sides  intersect  or 
meet. 

A  parallelogram  is  a  quadrilateral  whose  opposite- sides 
are  parallel.  Draw  one. 

A  trapezoid  is  a  quadrilateral  two  of  whose  opposite 
sides  only  are  parallel.  Draw  one. 

A  trapezium  is  a  quadrilateral  no  two  sides  of  which  are 
parallel.  Draw  one. 

The  altitude  of  a  parallelogram  is  the  distance  between 
either  pair  of  parallel  sides;  of  a  trapezoid,  it  is  the  distance 
between  its  parallel  sides.  Draw  the  altitude  of  each. 

A  rectangle  is  a  parallelogram  whose  angles  are  all  right 
angles.  A  rectangle  with  all  sides  equal  is  called  a  square. 
Draw  each. 

A  rhomboid  is  a  parallelogram  with  oblique  angles. 

A  rhombus  is  an  equilateral  rhomboid.     Draw  one. 


34  Plane  Geometry. 

Proposition  XXV. 

63.  The  diagonal  of  a  parallelogram  divides  it  into  two 
equivalent  triangles. 

Statement:  If  ABCD  is  a  O,  BD  is  a  diagonal  and  divides 
the  O  into  the  two  equivalent  triangles  ABD  and  DEC. 

Analysis:  Triangles  are  equal  if  they  have  two  sides  and 
included  angle;  or  two  angles  and  included  side,  or  three 
sides  equal  each  to  each. 


Fig.  31. 

We  may  choose  any  one  of  these  methods  of  comparison 
that  suits  our  data. 

Proof:  BD  is  common  to  both  triangles,  and  since,  by 
definition,  AD  is  ||  to  BC  and  AB  is  ||  to  DC,  and  each  of  these 
parallel  pairs  is  cut  by   the  transversal    DB,   Z  ABD  = 
Z  BDC  and  Z  ADB  -  Z  DEC.     What  follows? 

64.  Cor.     Since    the   triangles  ABD  =  triangle    BDC, 
their  homologous  parts  are  all  equal,   hence   the  parallel 
sides  of  a  parallelogram  are  also  equal. 

NOTE.  -  Two  figures  are  said  to  be  equivalent  if  they 
inclose  equal  space,  but  cannot  be  made  to  coincide;  equal, 
if  they  inclose  equal  space  and  will  also  coincide. 

Proposition  XXVI. 

65.  The  two  diagonals   of  a   parallelogram   bisect  each 
other. 


Plane  Geometry. 


35 


Statement:  ABCD  is  a  O;  then  its  diagonals  AC  and  DB 
bisect  each  other,  that  is,  E  is  the  middle  point  of  both. 
Analysis:  Same  as  Prop.  XXV. 


Fig.  32. 

Proof:  In  the  triangle  AEB  and  DEC,  AB  is  equal  and 
parallel  to  DC  by  last  proposition.  They  are  cut  by  the 
transversals  AC  and  DB. 

Complete  the  proof. 

Exercise:  State  and  prove  the  converse  of  above 
proposition. 

Proposition  XXVII. 

66.  //  two  sides  oj  a  quadrilateral  are  equal  and  parallel, 
the  figure  is  a  parallelogram. 


Fig.  33. 

Statement:  If  AB  and  CD  are  equal  and  parallel,  then 
ABDC  is  a  parallelogram. 

Analysis:  By  definition  a  parallelogram  has  parallel 
opposite  sides;  also  two  lines  are  parallel  if  they  do  not 


36  Plane  Geometry. 

meet,  or  if  a  transversal  makes  equal  alternate-interior  angles 
with  them.  It  is  clear  that  the  latter  criterion  must  be  used 
here.  It  remains  then  to  prove,  for  instance,  that  the  diago- 
nal makes  equal  angles  with  AC  and  BD,  which  can  be 
most  easily  established  by  proving  the  equality  of  the 
triangles  containing  them. 

Proof:  Draw  a  diagonal  CB;  then  in  the  two  triangles 
ABC  and  BCD,  AB  equals  CD,  and  since  also  AB  is  ||  to 
CD,  and  both  are  cut  by  CB,  the  alternate  angles  ABC  and 
BCD  are  equal;  also  CB  is  common,  /.  triangle  ABC 
equals  triangle  CBD.  (Why?) 

Hence,     Z  BCA  equals  Z  CBD,  and  /.  AC  is  ||  BD. 

Since  the  sides  are  parallel  in  pairs,  ABDC  is  a  parallelo- 
gram. 

Exercise:  Prove  that,  if  the  opposite  sides  of  a  quad- 
rilateral are  equal,  the  figure  is  a  parallelogram. 


Proposition  XXVIII. 

67.  //  three  or  more  parallels  intercept  equal  parts  of 
one  transversal,  they  intercept  equal  parts  of  every  trans- 
versal. 

Statement:  If  the  parallels  GF,  DK,  CL,  BM,  etc.,  inter- 
cept equal  parts  of  EM  as  FK  =  KL  =  LM,  etc.,  then  they 
intercept  equal  parts  on  any  other  line  AB,  as  DG  =  DC  = 
CB,  etc. 

Analysis:  To  compare  lines  or  geometrical  magnitudes 
of  any  sort  it  is,  as  a  general  rule,  necessary  to  connect  them 
or  their  equivalents.  If  we  draw  from  G,  D,  and  C,  lines 
||  to  EM,  we  evidently  get  the  equivalents  of  FK,  KL,  LM, 
etc.,  and  since  these  ||'s  are  connected  with  DG,  CD,  BC, 
etc.,  we  ought  to  be  able  to  derive  relations  between  them. 

Proof:  Draw  GX,  DY,  CZ,  etc.,  ||  to  EM. 


Plane  Geometry. 


37 


In  the  triangles  DGX  and  DYC,  GX  =  DY  (for  GX  = 
FK  and  DY  =  KL(Why  ?),  and  FK  =  KL  by  hypothesis), 


Fig.  34. 

Z  DGX  =  Z  CDY  (since  GX  is  ||  to  DY  and  they  are 
cut  by  AB),  Z  GDX  =  Z  DCY  (since  DK  is  ||  to  CL  and 
cut  by  AB),  hence  Z  GXD  =  Z  DYC.  "  (Why?) 

Then  we  have  two  angles  and  the  included  side,  .*.  triangle 
DGX  =  triangle  CDY,  and  hence  DG  =  CD;  likewise, 
CD  can  be  proved  equal  to  BC,  etc. 

68.  Cor.  I.  If  a  line  is  parallel  to  one  side  of  a  triangle; 
and  bisects  another  side,  it  E 

bisects  the  third  side  also. 
That  is,  in  the  triangle  AEB, 
if  DC  is  ||  to  AB  and  bisects 
AE  it  bisects  EB  also. 

Draw  an  auxiliary  line 
through  E  ||  to  DC  and  AB 
and  apply  Prop.  XXVIII.  Fig  35 

Complete  proof  (Fig.  35). 


38  Plane  Geometry. 

69.  Cor.  II.  The  line  joining  the  mid-points  of  two  sides 
of  a  triangle  is  parallel  to  the  third  side  and  is  one  half  the 
length  of  the  third  side  (Fig.  36). 


o 

Fig.  36. 

That  is,  if  M  and  N  are  the  mid-points,  respectively,  of 
xy  and  zy,  then  MN  is  ||  to  ocz  and  MN  =  ^  xz. 

First  part:  A  line  drawn  through  N  ||  to  xz  passes  through 
M  by  Cor.  I,  and  since  two  points  determine  a  straight 
line,  MN  must  be  this  line  which  is  parallel  to  xz. 

Second  part:  Draw  ON  through  N  ||  to  xy.  Then, 
since  MN  is  ||  to  xO  and  ON  is  ||  to  xM,  #MNO  is  a 
parallelogram;  hence  MN  =  xO,  but  O  is  the  mid-point 
of  xz.  (Why?) 

/.  MN  =  xO  =  J  xz. 

70.  Cor.  III.  The  median  of  a  trapezoid  is  parallel  to 
the  bases,  and  is  equal  to  half  the  sum  of  the  bases. 

That  is,  DC  is  ||  to  AB  and  FG  =  }  (AB  +  FG). 

First  part:  Draw  the  diagonal  AG  and  let  E  be  its  mid- 
point. 

Connect  E  with  D  and  C  by  straight  lines  DE  and  EC; 
since  AB  is  ||  to  FG,  DE  and  EC  are  both  ||  to  AB  and  FG, 
and  since  only  one  line  could  be  drawn  through  the  point 


Plane  Geometry.  39 

E  ||  to  either  AB  or  FG,  BE  and  EC  must  be  collinear,  that 
is,  form  parts  of  the  same  straight  line. 


Fig.  37. 

Second  part:   In  the  triangles  AFG  and  ABG,  by  Cor.  II, 

DE  =  J  FG, 

and  EC  =  i  AB. 

Add  ;  DE  +  EC  =  i  (FG  +  AB), 

or  DC  =  i  (FG  +  AB). 

POLYGONS. 

Definitions. 

71.   A  polygon  is  a  plane  figure  of  any  number  of  sides. 

The  vertices  of  a  polygon  are  the  points  where  the  sides 
intersect. 

A  diagonal  of  a  polygon  is  a  line  joining  any  two  vertices 
not  adjacent. 

A  polygon  is  said  to  be  equilateral  where  its  sides  are  all 
equal;  equangular,  when  its  angles  are  all  equal. 

A  polygon  is  said  to  be  convex  when  no  side  produced 

will  enter  the  polygon;  as,    |       | 

If  any  side  producing  does  enter  the  polygon  it  is  said 
to  be  concave;  as 


4O  Plane  Geometry. 

The  angle  within  a  concave  polygon,  formed  by  the  sides 
which  would  enter  the  polygon  if  produced,  is  called  a 
re-entrant  angle. 

Only  convex  polygons  are  considered  here.     The  mutu- 
ally equal  angles  in  two  or  more  equangular  polygons  are 
called   homologous  angles,   and  the   sides  including  these 
equal  angles  are  called  homologous  sides. 
A  polygon  of  three  sides  is  a  triangle. 
A  polygon  of  four  sides  is  a  quadrilateral. 
A  polygon  of  five  sides  is  a  pentagon. 
A  polygon  of  six  sides  is  a  hexagon. 
A  polygon  of  seven  sides  is  a  heptagon. 
A  polygon  of  eight  sides  is  an  octagon,  etc. 
This  may  be  represented  schematically  as  follows: 

triangle 3  sides. 

quadrilateral     ...  4  sides. 

pentagon 5  sides. 

hexagon 6  sides. 

heptagon 7  sides. 

octagon      .....  8  sides. 

nonagon 9  sides. 

decagon 10  sides,  etc. 

Proposition  XXIX. 

73.  The  sum  of  the  interior  angles  of  a  polygon  is  equal 
to  two  right  angles  taken  as  many  times  less  two,  as  the 
polygon  has  sides. 

Statement:  In  the  polygon  ABCDEF,  ZA+ZB+ 
Z  C  +  Z  D  +  Z  E  +  Z  F  =  two  right  angles  taken 
(6—2)  times,  that  is  =  eight  right  angles. 

Analysis:  Since  we  can  divide  the  polygon  into  triangles 
by  diagonals  from  any  vertex,  and  since  we  know  the  sum 


polygons. 


Plane  Geometry.  41 

of  the  angles  in  any  triangle,  it  is  desirable  to  make  this 
division. 

Proof:  Draw  diagonals  from  A  to  C,  D  and  E,  forming 
triangles  BAG,  CAD,  DAE,  EAF. 

Since  each  of  the  outside  triangles  ABC  and  AFE  use 
two  of  the  sides  of  the  polygon,  and  each  of  the  other  tri- 


Fig.  38. 

angles  one,  there  will  be  two  less  triangles  than  sides  of  the 
polygon. 

The  sum  of  the  angles  in  all  of  the  triangles  is  equal  to 
the  sum  of  the  angles  of  the  polygon. 

There  being  two  less  triangles  than  sides  of  the  polygon, 
there  will  be  two  right  angles  as  many  times  as  the  polygon 
has  sides,  less  two. 

In  general,  if  the  polygon  has  N  sides,  there  will  be  two 
right  angles  repeated  N  —  2  times. 


Proposition  XXX. 

74.  The  sum  of  the  exterior  angles  of  a  polygon  equals 
four  right  angles. 

Statement:  If  the  sides  of  the  polygon  ABCFH  are  pro- 
duced in  order,  the  sum  of  the  exterior  angles  ABE,  BCD, 
CFG,  FHK,  LAH  =  four  right  angles. 


42  Plane  Geometry. 

Analysis:  The  exterior  angles  and  their  corresponding 
interior  angles  at  each  vertex  equal  two  right  angles  and 
we  know  the  sum  of  the  interior  angles. 


Fig.  39- 

Proof:  As  an  illustration,  a  pentagon  is  taken,  but  what 
is  true  of  the  pentagon  is  evidently  true  of  any  other  polygon, 
as  to  angle-sums, 

EBA  +  ABC  =  two  right  angles. 

BCD  +  BCF  =  two  right  angles. 

CFG  +  CFH  =  two  right  angles. 

FHK  +  FHA  =  two  right  angles. 

HAL  +  HAB  —  two  right  angles. 

Add ;  (EBA  +  BCD  +  CFG  +  FHK  +  HAL)  +  ( ABC  + 
BCF  +  CFH  +  FHA  +  HAB)  -  10  rt.  angles. 

But  by  the  last  proposition, 

ABC  +  BCF  +  CFH  +  FHL  +  HAB  =  2  (5  -  2)  =  6  rt. 
angles. 

/.  FBA  +  BCD  +  CFG  +  FHK  +  HAL  =  4  right 
angles  =  (10  —  6)  right  angles. 

In  general,  if  there  are  n  sides,  the  sum  of  the  exterior 
angles  +  sum  of  the  interior  angles  =  n  X  2  rt.  angles  and 
sum  of  the  interior  angles  =  (n—  2)  2  rt.  A  =  (211  —  4)  rt.  A. 

.*.  Sum  of  exterior  angles  +  ('in  —  4)  rt.  A  =  2n  rt.  A. 

.*.  Sum  of  exterior  angles  =  \2n  —  (211  —  4)]  rt.  angles  = 
4  rt.  angles. 


Plane  Geometry  43 


EXERCISE  I. 

1.  Show  that  the  bisector  of  an  angle,  also  bisects  its 
..vertical  angle. 

2.  Show  that  the  bisectors  of  two  supplementary  adja- 
cent angles  are  perpendicular  to  each  other. 

3.  One  of  the  four  angles,  formed  by  the  intersection  of 
two  lines,  equal  |  of  a  right  angle.     Find  each  of  the  other  3. 

4.  The  sum  of  two  of  the  vertical  angles  formed  by  the 
intersection  of  two  lines,  is  3  times  the  sum  of  the  other  two. 
What  is  the  value  of  each  angle  ? 

5.  An  angle  is  5  times  its  supplementary  adjacent  angle. 
What  is  its  value? 

6.  Two  parallel  lines  are  cut  by  a  transversal;  prove 
that  the  alternate  exterior  angles  are  equal. 

7.  Prove  that  the  bisectors  of  the  two  alternate  interior 
angles  of  two  parallels  cut  by  a  transversal,  are  parallel. 

8.  The  bisector  of  the  exterior  angle  at  the  vertex  of  an 
isosceles  triangle  is  parallel  to  the  base. 

9.  Find  the  angles  A,  B,  and  C  of  a  triangle  if  A  =  26 
and  B  =  2C. 

10.  If  the  vertical  angle  of  an  isosceles  triangle  is  f  of  a 
right  angle,  show  that  the  triangle  is  equilateral. 

11.  Show  that  the  sum  of  the  lines  drawn  from  any  point 
inside  a  triangle  to  the  vertices  is  greater  than  half  the  peri- 
meter and  less  than  the  whole  perimeter. 

12.  Can  sides  of  the  length  7,  8,  and  17  inches  respec- 
tively form  a  triangle.     Give  reason  for  answer. 

13.  In  the  isosceles  triangle  xyz,  A  being  a  point  in  xz 
a  line   is  drawn    from  y  to  A  so  that   xA.  >   Az.     Prove 
Z.  zky  <  yAz. 

14.  In  an  isosceles  triangle  show  that  the  bisectors  of  the 
base  angles,  form  an  isosceles  triangle  with  the  base. 


44  Plane  Geometry. 

15.  If  one  angle  of  a  parallelogram  is  f  of  a  right  angle, 
what  are  the  ether  angles? 

16.  If  two  sides  of  a  quadrilateral  are  equal  and  two  of 
the  respectively  opposite  angles  are  equal,  it  is  a  parallelo- 
gram. 

17.  Any  line  drawn  through  the  intersection  point  of  the 
diagonals  of  a  parallelogram  and  terminating  in  opposite 
sides  is  bisected  by  the  point. 

18.  In  an  isosceles  trapezoid  the  angles  at  the  extremi- 
ties of  a  base  are  equal. 

19.  The  line  joining  two  points  on  opposite  sides  of  a 
given  line  and  equally  distant  from  it,  is  bisected  by  the 
given  line. 

20.  What  is  the  size  of  each  interior  angle  of  a  regular 
decagon  ? 

21.  One  interior  angle  of  a  regular  polygon  is  if  right 
angles,  how  many  sides  has  the  polygon  ? 

22.  In  laying  a  tile  floor,  what  different  kinds  of  regular 

polygons  can  be  used,  and  how  many 
of  each  kind  will  be  required  around 
any  one  point  ? 

23.  The  lines   joining    the    middle 
points  of  the  opposite  sides  of  a  quad- 
rilateral bisect  each  other. 

24.  If  from  any  point  in  the  base 

of  an  isosceles  triangle  (as  E,  in  ACF)  two  lines  be  drawn 
parallel    respectively  to  the  equal  sides,  the  perimeter  of 
the   quadrilateral    (BEDC)    thus  formed  equals   the  sum 
of  the  equal  sides  (AC  and  CF)  and  hence  is  constant. 

25.  How  many  sides  has  a  polygon,  if  the  sum  of  its 
interior  angles  exceeds  the  sum  of  its  exterior  angles  by 
10  right  angles? 


Plane  Geometry.  45 

26.  The  middle  point  of  the  hypotenuse  of  a  right  tri- 
angle is  equally  distant  from  the  three  vertices. 

27.  Show  how  to  determine  a  point  in  a  straight  line, 
the  sum  of  whose  distances  from  two  fixed  points  on  the 
same  side  of  the  line  (and  unequally  distant  from  it)  shall 
be  the  least  possible. 

28.  Show  that  the  lines  in  27  joining  the  two  exterior 
points  with  the  point  in  the  line,  make  equal  angles  with 
the  line. 

29.  If  a  line,  drawn  through  the  vertex  of  a  triangle,  par- 
allel to  its  base,  bisects  the  exterior  angle,  the  triangle  is 
isosceles. 

30.  The   angle,   formed   by   the   bisectors   of   any   two 
angles    of    an    equilateral    triangle,   is   double   the   other 
angle. 

31.  If  the  diagonals  of  a  parallelogram  are  perpendicu- 
lar to  each  other  and  equal,  the  figure  is  a  square. 

32.  The  angle  formed  by  the  bisectors  of  the  interior 
angle  between  the  base  and  one  side  of  a  triangle,  and  of 
the   exterior   angle   between   the   base   produced   and   the 
other  side,  is  one  half  the  vertical  angle  of  the  triangle. 

THE    CIRCLE. 

Definitions. 

74.  A  circle  is  a  plane  figure  whose  boundary  is  a 
curved  line,  every  point  of  which  is  equally  distant  from  a 
point  within,  called  its  center. 

The  boundary  curve  is  called  the  circumference. 

75.  The  radius  is  a  straight  line  joining  the  center  to 
any  point  on  the  circumference. 

The  word  radius  often  refers  to  the  length  of  this  line, 
not  necessarily  to  the  line  itself. 


46  Plane  Geometry. 

76.  The  diameter  is  a  straight  line  through  the  centre 
joining  any  two  opposite  points  on  the  circumference. 

Necessarily  all  radii  and  all  diameters  of  the  same  circle 
are  equal,  and  the  diameter  is  twice  the  radius. 

77.  An  arc  is  any  part  of  the  circumference. 

A  chord  is  a  straight  line  joining  the  ends  of  an  arc. 
A  chord  is  said  to  subtend  its  arc. 

Every  chord  subtends  two  arcs  from  the  nature  of  a 
circle.     The  smaller  one  is  the  one  usually  referred  to. 

78.  A  polygon  is  said  to   be  inscribed  in  a  circle  when 
its  vertices  are  points  in  the  circumference.     The  circle  is 
said  to  be  circumscribed  about  the  polygon. 

79.  An  angle  is  said  to  be  inscribed  in  a  circle  when  its 
vertex  is  in  the  circumference  and  its  sides  are  chords. 

80.  A  segment  of  a  circle  is  the  portion  of  it  bounded  by 
any  arc  and  its  chord. 

81.  A  sector  of  a  circle  is  the  portion  of  it  bounded  by 
any  two  radii  and  the  arc  between  their  extremities. 

Represent  a  circle  with  all  the  parts  denned. 

82.  Clearly  any  diameter  divides  a  circle  into  two  equal 
parts  as  also  its  circumference. 

Also  circles  having  equal  radii  are  equal. 


Proposition  I. 

83.   A  diameter  is  the  greatest  chord  of  any  circle. 

Statement:  Let  AB  be  a  diameter  of  the  circle  O,  then  is 
AB  greater  than  any  other  chord,  say  CD. 

Analysis:  Since  all  diameters  are  equal,  we  can  draw 
another  diameter  through  either  end  of  the  chord  and  thus 
get  a  comparison,  remembering  that  a  diameter  is  twice  a 
radius. 


Plane  Geometry. 


47 


Fig.  40. 

Proof :  Draw  another  diameter  DE  through  D ;  also  join 
C  and  O.    Then  DO  +  CO  >  CD,  but    DO  +  CO  = 
DE  (Why?)  and  DE  =  AB,  /.  AB  >  CD  any  other  chord. 

Proposition  II. 

84.   In   the  same  circle  or  in  equal  circles,  equal  angles 
at  the  centre  intercept  equal  parts  of  the  circumference  (equal 

arcs}. 

Statement:  Let  A  and  B  be  two  equal  circles  and  Z_  DAC 
=  Z  FBG,  then  arc  DC  =  arc  FG. 

Analysis:  Superposition. 


Fig.  41- 

Proof:  Place  circle  B  on  circle  A,  making  BG  coincide 
with  AC;  since  the  radii  are  equal  G  falls  on  C,  and  since 


48  Plane  Geometry. 

Z  DAC  =  Z  GBF,  BF  falls  on  AD,  and  BF  =  AD,  .-.  F 
falls  on  D,  hence  arc  FG  coincides  with  arc  DC.      (Why?) 
Prove  the  converse  of  Prop.  II. 

Definitions. 

85.  The   last    proposition    would    suggest   that    an    arc 
would*  make  a  convenient  measure  for  an  angle.     Since 
four  right  angles  is  the  aggregate  of  all  angles  formed  about 
a  point,  no  more,  no  less,  the  sum  of  angles  formed  at  the 
centre  of  every  circle  is  the  same,  hence  every  circumfer- 
ence, if  used  as  a  measure,  must  contain  the  same  number 
of  units. 

It  has  been  agreed  that  ever}'  circumference  shall  be  said 
to  contain  360  of  these  units,  called  degrees.  As  the  circles 
vary  in  size,  plainly  the  lengths  of  these  degrees  vary  with 
the  circle  ;  hence  it  is  clearly  necessary,  in  comparing  angles, 
to  employ  parts  of  the  same  circumference  or  of  circum- 
ferences having  equal  radii. 

86.  Each  degree  is  arbitrarily  divided  into  60  minutes, 
and  each  minute  into  60  seconds.     Compare  the  method  of 
determining  units  of  linear  measure. 

87.  A  secant  line  is  any  line  which  cuts  the  circle;  or  a 
line  having  two  points  in  common  with  the  circle. 

88.  A  tangent  is  a  line  touching  the  circle  in  only  one 
point;    or    a    line   having    one    point    in    common   with 
the  circle. 

Proposition  III. 

89.  In  the  same  circle  or  equal  circles,   equal  arcs  are 
subtended  by  equal  chords ;  and  of  two  unequal  arcs  the 
greater  is  subtended  by  the  greater  chord. 

Statement:  If  M  and  N  are  two  equal  circles,  and  the  arcs 


Plane  Geometry. 


49 


AB  and  OR  are  equal,  then  chord  AB  =  chord  OR;  and  if 
arc  SR  >  arc  AB,  chord  SR  >  chord  AB. 

Analysis:  Since  these  chords  can  easily  be  incorporated 
into  triangles  their  comparison  is  simple. 


Fig.  42. 

Proof:  Draw  radii  MR,  MO,  and  MS,  also  chords  OR 
and  OS  in  M,  and  NB  and  NA,  also  chord  AB  in  N.  In 
triangles  MOR  and  NAB, 

MO  =  NA 
MR  =  NB. 

Z  OMR  -  Z  ANB  (Why?),  /.  triangle  MOR  equals 
triangle  NAB  and  .'.  chord  OR  =  chord  AB. 

Again  in  triangles  RMS  and  OMR,  SM  =  OM  and  MR 
is  common  but  Z  RMS  >  Z  OMR  (by  hypothesis), 

.-.  SR  >  OR  (Why?) 

or  SR  >  AB  -  OR. 


EXERCISE. 

State  and  prove  the  converse  of  Prop.  III. 


50  Plane  Geometry. 

Proposition  IV. 

90.  A  diameter  perpendicular  to  a  chord  bisects  the  chord 
and  the  arc  it  subtends. 

Statement:  Let  A  be  any  circle  and  BD  a  chord  in  it, 
then  the  diameter  FE  (or  the  radius  AE)  J_  to  BD  bisects 
BD  at  C  and  also  the  arc  BED  at  E. 


Analysis:  The  proof  that  C  is  the  middle  point  of  BD, 
that  is,  BC  =  CD,  is  evidently  dependent  upon  the  equality 
of  the  right  triangles  BAG  and  CAD. 

Proof:  Draw  the  radii  AB  and  AD,  forming  with  AC  and 
BD  two  right  triangles,  BAG  and  CAD,  wherein  AB  =  AD 
and  AC  is  common,    .'.  BC  -  CD   and  also   Z  BAG  = 
Z  CAD,  hence  arc  BE   =  arc  BD. 

What  might  be  inferred  as  corollaries? 

Proposition  V. 

91.  In  the  same  circle  (or  in  equal  circles}  equal  chords 
are  equally  distant  from  the  centre. 

Statement:  In  the  circle  M,  the  chord  OS  =  chord  PR, 
then  NM  =  MQ. 


Plane  Geometry.  51 


Fig.  44. 

Analysis:   Form  right  triangles  with  radii. 

Proof:  Draw  radii  MO  and  MP.  Then  in  the  right 
triangles  MON  and  PMQ,  OM  =  MP,  and  ON  =  PQ 
(Why?),  .'.  MN=  MQ. 

State  and  prove  the  converse  of  Prop.  V. 


Proposition  VI. 

92.  In  the  same  circle  (or  in  equal  circles)  unequal  chords 
are  unequally  distant  from  the  centre  and  the  greater  chord 
is  at  the  lesser  distance. 

Statement:  If  in  the  circle  D,  the  chord  EG  >  chord  AC, 
BD  >  FD. 

Analysis:  If  we  can  include  these  lines  BD  and  FD  in 
the  same  triangle,  they  can  readily  be  compared.  By  Prop. 
V,  the  chord  EG  and  its  distance  from  the  centre  FD  can 
be  duplicated  by  a  chord  drawn  from  C  or  A,  and  hence  a 
connection  established  with  AC. 

Proof:  Draw  CK  =  EG,  also  DL  perpendicular  to  CK. 
Then  by  Prop.  V,  DL  =  FD  connect  B  and  L. '  Then 
in  the  triangle  BCL,  thus  formed,  CL  >  BC,  for 
CL  =  }  CK  and  BC  =  i  AC  (Why?)  and  CK  >  AC, 


K 
Fig.  45- 

hence  Z  CBL  >  Z  CLB  (Why?).  Z  CBD  =  Z  CLD, 
(both  right  angles). 

Subtract  Z  CBL  >  Z  CLB. 

ZCBD- Z  CBL=  Z  LBD  <  ZBLD-Z  CLD -ZCLB, 
since  the  greater  of  two  quantities  leaves  the  less  remainder 
when  they  are  substracted  from  the  same  or  equal  .quantities. 

But  if  ZLBD  <  ZBLD,  DL  <  BD  (why?),  or  since 
DL  =  DF,  DF  <  BD. 

EXERCISE. 

State  the  converse  of  Prop.  VI,  and  indicate  the  method 

of  proof. 

Proposition  VII. 

93.  A  tangent  to  a  circle  is  perpendicular  to  the  radius 
drawn  to  the  point  of  tangency. 

Statement:  If  FE  is  tangent  to  the  circle  A  at  B,  then  it 
is  perpendicular  to  AB. 

Analysis:  The  perpendicular  is  the  shortest  line  from  a 
point  to  a  line. 

Proof:  Draw  AD  any  other  line  from  A  to  FE,  cutting 
the  circle  at  C.  Since  by  definition  the  tangent  to  a  circle 
touches  it  in  but  one  point  the  point  D  must  lie  outside  the 
circle,  hence  AD  >  AC,  but  AC  =  AB. 

V.  AD  >  AB;  hence  AB  is  shorter  than  any  other  line 
from  A  to  FE.  .'.  AB  is  perpendicular  to  FE. 


Plane  Geometry. 


53 


Fig.  46. 

94.  Cor.  I.   A  straight  line  perpendicular  to  a  radius  at 
its  extremity,  is  tangent  to  the  circle. 

95.  Cor.  II.   A  perpendicular  to  a  tangent  at  the  point  of 
contact,  passes  through  the  centre  of  the  circle. 

Proposition  VIII.     Construction. 

96.  To  draw  a  circumference  through  any  three  points 
not  in  the  same  straight  line. 

Statement:  Let  A,  B  and  C  be  any 
three  points  not  in  the  same  straight 
line;  to  pass  a  circumference  through 
them. 

Analysis:  Since  every  point  of  a  cir- 
cumference   must    be    equally  distant 
from  its  centre,  a  point  must  be  found 
equally  distant  from  A,  B  and  C,  which 
immediately  suggests  the  perpendicular  bisector. 

Proof:  Join  A,  B  and  C  by  the  straight  lines  AB  and  BC. 
Erect  to  them  the  perpendicular  bisectors  DF  and  DE 
respectively.  Now  DF,  being  the  perpendicular  bisector  of 
AB,  contains  all  points  equally  distant  from  A  and  B,  and 
DE,  for  a  like  reason,  contains  all  points  equally  distant 
from  B  and  C;  hence  the  point  where  DF  and  DE  meet 
(why  do  they  meet?)  will  then  be  equally  distant  from  A,  B 


54  Plane  Geometry. 

and  C,  because  it  is  a  point  common  to  both  perpendicular 
bisectors  and  possesses  all  the  characteristics  of  both. 
Hence  with  the  point  of  intersection,  D,  as  a  centre  and  a 
radius  DB,  a  circumference  will  pass  through  A,  B  and  C. 

It  is  the  only  circumference  that  can  pass  through  A,  B 
and  C  because  two  lines  (the  perpendicular  bisectors  in 
this  case)  can  intersect  in  only  one  point. 

97.  Cor.  Two  circumferences  cannot  intersect  in  more 
than  two  points.  (Why?) 

Proposition  IX. 


Fig.  48. 

98.  The  tangents  to  a  circle  from  an  external  point  are 
equal  in  length,  and  make  equal  angles,  with  the  line  joining 
the  point  to  the  centre. 

Statement:  The  tangents  DB  and  DC  drawn  from  the 
point  D  to  the  circle  A  are  equal  and  Z  BDA  =  Z  ADC. 

Analysis:  Comparison  of  two  right  triangles.  Complete 
the  proof. 


Plane  Geometry. 


55 


Proposition  X. 

99.  //  two  circles  intersect,  the  line  of  centers  is  perpen- 
dicular to  their  common  chord  at  its  middle  point. 

Statement:  Let  the 
circles  M  and  N  in- 
tersect, with  common 
chord  OP,  then  MN 
is  _L  to  OP  at  its 
middle  point  O'. 

Analysis :  The  con- 
ditions  for  perpendi- 
cular bisector  must  be 

Fig.  49. 

established. 

Proof:  Join  O  and  P  with  both  M  and  N.  Then  ON  = 
NP  and  OM  =  MP,  hence  M  and  N  are  each  equally 
distant  from  O  and  P,  hence  they  are  points  on  perpen- 
dicular bisector. 

100.  Cor.     If  two  circles  are  tangent,  the  line  of  centers 
passes  through  the  point  of  contact.  (Why?) 

Proposition  XI.     Construction. 

101.  To  find  the  ratio  of  two  given  lines. 
Let  the  lines  be  AB  and  CD. 


Fig.  50. 


Lay  off  CD  on  AB  as  many  times  as  possible;  in  this 
case,  twice  with  a  remainder  EB. 


Plane  Geometry. 


^         off  EB  on  CD  as  many  times  as  possible;  here  twice, 
with  a  remainder  GD. 

Lay  off  GD  on  EB  as  many  times  as  possible;  here  once, 
with  remainder  KB. 

Lay  off  KB  on   GD  as  many  times  as  possible;  here 
twice,  with  remainder  LD. 
LD  goes  into  KB  exactly  twice,  say. 
Then 

KB  =  2LD  (i). 

GD  =  GL  +  LD  =  2  KB  +  LD  =  5  LD  (from  (i))  (2) 
EB  =  EK  +  KB  =  GD  +  KB  =  7  LD    .     .     .   (3) 

from  ((i)  and  (2)) 
CD  =  CG  +  GD  =  2  EB  +  GD  =  19  LD       .     .   (4) 

from  ((2)  and  (3)) 
AB  =  AE  +   EB  =  2  CD  +  EB  =  45  LD       .     .   (5) 

from  ((3)  and  (4)) 
.  CD  =  19  LD  =  i£_ 
"AB       45LD~45' 

By  the  same  process,  plainly,  any  two  lines,  straight, 
or  of  the  same  curvature,  may  be  compared. 

Proposition  XII. 

102.   In  the  same  circle  or  in  equal  circles,  two  central 
angles  have  the  same  ratio  as  their  intercepted  arcs. 


Fig.  51. 


Plane  Geometry.  57 

Statement:  Let  O  and  C  be  equal  circles  having  central 
angles  BOA  and  ECD  commensurable;  also  BOA  and  FCG 
incommensurable. 

ZBOA       arcAB        ,  ZBOA       arcAB 

ZECD  =  aTc-ED  and  ZFCG  =  ScTo * 

Analysis:  A  case  for  common  measure,  clearly. 
Proof:    Case    I.     Arcs    AB    and    ED    commensurable. 
Take  any  common  measure,  as  small    arc  x,  with  corre- 
sponding central  angle,  in  an  equal  circle. 

Say  it  is  contained  in  AB,  m  times,  and  in  ED,  n  times, 
then 

arc  AB      m 
arc  ED  =  ^" 

Draw  radii  in  each  circle  to  the  points  of  division,  divid- 
ing the  angle  AOB  into  m  parts,  and  the  angle  ECD  into 
n  parts,  all  equal.  (Why?) 

ZAOB      m 
ZECD  =  n 
combining  these  two  proportions,  . 

ZAOB  _  arc  AB 
ZECD  ~  arc  ED* 

Case  II.  Arcs  AB  and  EG,  incommensurable.  Apply 
the  same  unit  of  measure  x  to  ACB  and  FCG;  say,  it  is 
contained  in  x\OB  m  times  (as  before)  and  leaves  a  re- 
mainder in  FCG,  represented  by  KCF. 

Now  the  arcs  AB  and  KG  are  commensurable,  for  both 
contain  the  common  measure,  hence  by  Case  I, 
ZAOB     _  arc  AB 
ZKCG  -    arc  KG  ' 

In  division  the  remainder  must  always  be  less  than  the 
divisor,  hence  the  remainder-arc  KF  is  less  than  the  unit 
used. 


Plane  Geometry. 


If  a  smaller  unit  is  employed  to  compare  the  two  arcs 
(and  angles),  there  will  be  a  smaller  remainder. 

Suppose  a  unit  is  chosen  (contained  exactly  in  AB)  so 
small  that  the  remainder  is  too  small  to  be  even  imagined; 
although  the  point  K  representing  the  initial  point  of  the 
remainder  —  arc  FK  is  so  near  F  that  this  remainder- 
arc  is  inconceivably  small,  still  the  proportion 

Z  AOB       arc  AB      . 

—-—7  holds  true,  because  AB    and   KG  both 
Z  KCG      arc  KG 

contain  this  very  small  unit  an  integral  number  of  times. 
Since  by  decreasing  the  unit  (which  we  can  do  at  will) 
we  can  bring  the  point  K  as  near  to  F  as  we  please,  and 
the  above  proportion  will  continue  always  to  be  true,  it 
will  hold  also  at  F. 

.  Z  AOB  =  arc  AB 

"ZFCG~arcFG" 

Proposition  XIII. 

103.  An  inscribed  angle  is  measured  by  half  the  arc 
intercepted  by  it. 

A 


D 
Fig.  52  b. 

Statement:  Let  BAG  be  an  angle  inscribed  in  circle  O, 
intercepting  arc  BC,  to  prove  Z  BAG  =  J  arc  BC. 


Plane  Geometry.  59 

Analysis:  Since  we  know  that  a  central  angle  is  measured 
by  its  arc,  it  is  necessary  to  express  this  inscribed  angle  in 
terms  of  a  central  angle  (or  central  angles)  or  of  its  arc. 

Proof:  Draw  the  diameter  AD  from  the  vertex  A,  also 
the  radii  OC  and  OB  to  points  C  and  B.  Triangles  AOB 
and  AOC  are  isosceles.  (Why?) 

/.  Z  OAB  =  Z  OBA  and  Z  OAC  =  Z  OCA. 

Also    Z  DOB  is  the  exterior  angle  of  the  triangle  OBA. 
hence,     Z  DOB  =  Z  OAB  +  Z  OBA  =  2  OAB. 

For  same  reason  Z  DOC  =  Z  OAC  +  ZOCA-2  OAC. 

That  is,  Z  OAC  =  J  Z  DOC 

and  Z  OAB  =  J  Z  DOB 

Z  BAG  =  J  Z  BOC 

subtracting  if  diameter  AD  is  outside  the  angle;  adding 
if  it  is  within.     (See  Fig.  52  b  and  Fig.  52  a.) 

But  J  ZBOC=  iarcBC.    '        (Why?) 

.'.  ZBAC  =  JarcBC. 

104.  Cor.  I.   An   angle   inscribed  in   a   semicircle   is   a 
right  angle. 

105.  Cor.  II.   An  angle  is  acute  or  obtuse  according  as  it 
is  inscribed  in  a  segment  greater  or  less  than  a  semi-circle. 


Proposition  XIV. 

106.  An  angle  formed  by  the  intersection  of  two  chords 
of  a  circle  is  measured  by  half  the  sum  of  the  intercepted 
arcs. 

Statement:  If  AC  and  BD,  two  chords  of  the  circle  O, 
intersect  at  E,  the  Z  BEC  (or  its  equal  AED)  is  measured 
by  i  (AD  +  BC);  and  Z  AEB  (or  its  equal  DEC)  is 
measured  by  J  (AB  +  CD). 

Analysis:  Since  we  have  learned  the  relation  between 
central  angles  and  their  arcs;  also  between  inscribed  angles 


6o 


Plane  Geometry. 
A 


Fig.  53- 

and  their  arcs,  it  is  clearly  desirable  to  express  the  angles 
under  discussion  in  terms  of  either  an  inscribed  or  of  a 
central  angle.  To  that  end  a  line  drawn  from  either  extrem- 
ity of  either  chord,  parallel  to  the  other,  will  form  a  related 
inscribed  angle. 

Proof:  From  C  draw  the  chord  CF  ||  to  BD,  then  by 
Prop.  XIII  Z  ACF  is  measured  by  i  arc  ABF,  and  Z  ACF 
=  ZAEB  (Why?).  .-.  Z  AEB  =  i  arc  ABF,  but  arc 
ABF  =  arc  AB  +  arc  BF  =  arc  AB  +  arc  CD  (arc  BF  = 
arc  CD,  being  between  parallel  chords),  hence  Z  AEt)  = 
i  (arc  AB  +  arc  CD). 

Proposition  XV. 

107.  An  angle,  included  by  a  tangent  and  a  chord  drawn 
from  the  point  of  contact,  is  measured  by  half  the  intercepted 
arc. 


Plane  Geometry.  61 

Statement:  Let  CD  be  a  tangent  to  the  circle  O,  meeting 
the  chord  AB  at  B,  the  point  of  tangency,  then  ABC  is 
measured  by  J  arc  AB. 

Analysis:  An  inscribed  angle  is  again  the  means  of  com- 
parison. 

Proof:  Draw  AE  ||  to  CD,  then  Z  EAB  =  J  EB  (=  i  AB) 
by  Prop.  XIII,  and  Z  EAB  =  Z  ABC.  .',  Z  ABC  -.  i  AB. 


Proposition  XVI. 

108.  An  angle  formed  by  two  secants,  two  tangents,  or  a 
tangent  and  a  secant,  drawn  to  a  circle  from  an  exterior 
point,  is  measured  by  half  the  difference  of  the  intercepted 
acrs.  The  analysis  is  the  same  as  that  for  Props.  XIV  and 
XV.  Prove  the  Proposition. 


EXERCISE  II. 

1.  A  diameter  of  a  circle  and  a  chord  are  produced  out- 
side the  circle  to  meet.     Show  that  the  part  of  the  diameter 
lying  outside  is  less  than  the  part  of  the  chord  outside. 

2.  If  the  circumference  of  a  circle  is  divided  into  four 
equal  parts,  the  chords  joining  these  points  form  a  rhombus 
or  a  square. 

3.  If  a  line  joining  the  middle  points  of  two  chords  is 
a  part  of  a  diameter,  the  chords  are  parallel. 

4.  A  chord  of  the  larger  of  two  unequal  concentric  cir- 
cles, that  is  tangent  to  the  inner  circle,  is  bisected  at  the 
point  of  tangency. 

5.  The  shortest  chord  that  can  be  drawn  in   a  circle 
through  a  given  point  in  it  is  perpendicular  to  line  joining 
that  point  and  the  centre. 


62  Plane  Geometry. 

6.  Two  circles  are  tangent  internally,  and  chords  of  the 
larger  circle  are  drawn  from  the  point  of  tangency  through 
the  extremities  of  a  diameter  of  the  smaller  circle.     Show 
that  the  line  joining  the  ends  of  these  chords  is  a  diameter 
of  the  larger  circle. 

7.  The  circumference  of  a  circle  is  divided  into  5  parts, 
such  that  one  part  is  twice  the  second  part,  3  times  the 
third,  4  times  the  fourth,  and  5  times  the  fifth.     What  is  the 
length  (in  degrees)  of  each  arc? 

8.  If  two  circles  are  tangent,  the  tangents  drawn  at  the 
ends  of  two  secants,  through  the  contact  point,  are  parallel. 

9.  Three  consecutive  sides  of  an  inscribed  quadrilateral 
subtend  arcs  of  82°,  99°,  67°  respectively.     Find  each  angle 
of  the  quadrilateral  in  degrees. 

10.  If  two  chords  intersect  at  right  angles  within  a  circle, 
the  sum  of  the  opposite  arcs  intercepted  equals  a  semi-cir- 
cumference. 

11.  Two  intersecting  chords  making  equal  angles  with 
the  diameter  through  their  point  of  intersection  are  equal. 

12.  The  sum  of  two  opposite  sides  of  a  circumscribed 
quadrilateral  is  equal  to  the  sum  of  the  other  two  sides. 

13.  The   straight   line   joining   the    middle   of    the   two 
non-parallel  sides  of  a  circumscribed  trapezoid  equals  J  its 
perimeter. 

14.  The  bisector  of  the  angle  formed  by  two  tangents  to 
a  circle  passes  through  the  centre. 

15.  The  circle  described  on  one  of  the  legs  of  an  isosceles 
triangle  as  diameter  bisects  the  base. 

1 6.  If  a  perpendicular  be  dropped  from  the  centre  of  the 
circumscribed  circle  upon  the  base  of  a  triangle,  the  angle 


Plane  Geometry.  63 

between  this  perpendicular  and  a  radius  to  the  extremity  of 
the  base  equals  the  vertical  angle. 

17.  What  is  the  locus  of  the  middle  points  of  a  system 
of  equal  chords. 

1 8.  If  a  straight  line  be  drawn  through  the  point  of  con- 
tact of  two  circles,  tangent  externally,  ending  in  the  circum- 
ferences, the  radii  drawn  to  its  extremities  are  parallel. 

19.  The  sum  of  the  angles  subtended  at  the  centre  of  a 
circle  by  two  opposite  sides  of  a  circumscribed  quadrilateral 
is  equal  to  two  right  angles. 

20.  If  a  circle  be  inscribed  in  a  right  triangle  the  sum  of 
its  diameter  and  the  hypotenuse  is  equal  to  the  sum  of  the 
legs. 

21.  If  two  circles  intersect  and  diameters  be  drawn  in 
each  circle  from  one  of  the  intersection  points,  a  line  joining 
the  other  ends  of  these  diameters  passes  through  the  second 
point  of  intersection. 

22.  The  angle  subtended  by  a  railroad  curve  at  the  cen- 
tre is  1 8°.     What  is  the  angle  between  the  tangent  at  one 
extremity  and  the  chord? 

23.  What  is  the  value  of  each  angle  of  a  regular  inscribed 
polygon  of  24  sides? 

24.  What  is  the  size  of  an  angle  of  a  regular  circum- 
scribed polygon  of  15  sides? 

25.  If   two   adjacent   sides   of   a   quadrilateral   subtend 
angles  of  67°  and  115°  respectively,  and  the  acute  angle 
between  the  diagonals  is  79°,  find  each  angle  of  the  quad- 
rilateral. 

26.  If  the  inscribed  and  circumscribed  circles  of  a  tri- 
angle are  concentric,  prove  the  triangle  equilateral. 


64 


Plane  Geometry. 


Problem  1. 


109.    To  construct  a  perpendicular  to  a  given  line  at  a 
given  point. 


v- 
/\ 


Fig.  55- 

Case  I:  Let  AB  be  the  given  line  and  C  the  given  point. 
Construction:  With  C  as  centre  and  any  convenient 
radius,  cut  the  line  AB  at  D  and  E  on  opposite  sides  of  C, 
with  D  and  E  successively  as  centres  and  any  radius  greater 
than  DC  (or  CE),  describe  two  arcs  intersecting  at  F;  a 
line  through  F  and  C  is  the  required  perpendicular.  The 
first  arcs  described,  cutting  AB  at  D  and  E,  are  plainly  for 
the  purpose  of  locating  two  points  (D  and  E)  equally  dis- 
tant from  C  on  AB,  from  which  the  two  arcs  (of  equal  radii) 
intersecting  at  F  give  a  point  equally  distant  from  D  and  E. 
Hence  we  have  the  conditions  for  a  perpendicular  bisector 
for  DE,  which  is  part  of  AB. 

Case  II:  If  the  point 
is  at  the  end  of  the  line, 
say  at  B,  take  any  point 
D,  outside  of  AB  and  not 
immediately  above  B ; 
with  DB  as  a  radius  (so 
that  the  arc  will  pass 
through  B),  describe  an 
arc  cutting  AB  at  C;  through  C  and  D  draw  a  straight 


Plane  Geometry.  65 

line,  prolonging  it  to  meet  the  circumference  again  at  E; 
the  line  joining  E  and  B  will  be  _L  to  AB  at  B,  for  the 
angle  EBC  is  inscribed  in  a  semi-circle.  The  reasons 
for  the  procedure  are  evident. 

Problem  2. 

110.   To  construct  a  perpendicular  to  a  given  line  from  a 
given  external  point. 


\ 


M- 


Fig.  57- 

Let  MN  be  the  given  line  and  O  the  given  point.  With 
O  as  a  centre,  and  a  radius  greater  than  the  distance  from  O 
to  MN,  describe  an  arc  cutting  MN  in  P  and  Q,  with  a 
larger  or  smaller  radius  (but  greater  than  J  PQ),  and  P  and 
Q  as  successive  centres,  describe  arcs  intersecting  at  R. 
The  line  through  R  and  O  will  be  the  required  _L 

It  is  clearly  necessary  tp  fix  two  points  in  MN  equally 
distant  from  O,  and  then  to  determine  another  point  outside 
MN  equally  distant  from  these  two  points. 

111.  Cor.   The  bisection  of  a  line  is  simply  a  corollary 
to  the  problem. 

Problem  3. 

112.  To  bisect  a  given  arc. 

This  is  easily  reduced  to  the  bisection  of  a  straight  line 
by  drawing  the  chord  of  the  given  arc,  XY. 


66 


Plane  Geometry. 


Then  taking  X  and  Y  successively  as  centres  and  a  radius 
greater  than  J  XY,  describe  two  arcs  intersecting  above  and 


two  intersecting  below  XY.  The  line  joining  these  inter- 
sections will  be  a  perpendicular  bisector  of  the  chord  XY 
and  hence  of  the  arc  XY. 

Problem  4. 

113.    To  bisect  a  given  angle. 

Let  A  be  the  given  angle. 

It  is  only  necessary  to  bear  in  mind  that  the  bisector  of  an 
arc  also  bisects  the  central  angle,  subtended  by  it,  and  hence 
the  problem  is  practically  the  same  as  the  last. 


Fig.  59. 

With  A  as  a  centre  and  any  radius,  describe  an  arc  inter- 
secting the  sides  of  the  angle  in  C  and  B.     (Why  is  this 


Plane  Geometry.  67 

done?)  With  C  and  B  successively  as  centres  and  a  radius 
greater  than  J  CB,  describe  two  arcs  intersecting  in  D,  the 
straight  line  joining  D  and  A  will  bisect  Z  A.  (Why?) 

Problem  5. 

114.   At  a  given  point  on  a  given  straight  line,  to  con- 
struct an  angle  equal  to  a.  given  angle. 


Fig.  60. 

Let  MN  be  the  given  line  and  Z  A  the  given  angle;  to 
construct  an  angle  at  O  on  MN  equal  to  Z  A. 

Analysis:  Since  angles  are  measured  by  the  arcs  that 
subtend  them  in  equal  circles  (Why  equal  circles?),  it  is 
necessary  to  select  a  radius  first,  that  is  to  be  used  in 
describing  the  arcs,  by  which  comparison  is  made,  and  then 
with  the  vertex  of  A  and  O  successively  as  centres,  to  describe 
arcs  which  are  to  be  used  in  comparing  the  angles. 

Take  any  radius,  and  with  the  vertices  A  and  O  succes- 
sively as  centres  describe  arcs.  The  arc  with  centre  A  will 
cut  the  two  sides  of  the  angle  at  B  and  C  with  a  radius 
equal  to  the  chord  of  the  arc  BC  and  R,  where  the  arc  with 
O  as  a  centre  cuts  MN  as  a  centre  cut  this  latter  arc  at  P, 
then  arc  PR  -  arc  BC.  (Why?) 

Hence  the  Z  POR  -  Z  A.  (Why?) 


68  Plane  Geometry. 

Problem  6. 

115.  To  draw  through  a  given  external  point  a  line 
parallel  to  a  given  line. 

Let  O  be  the  given  point,  and  MN  the  given  line;  to  draw 
through  O  a  line  parallel  to  MN. 


p\ 

B 
Fig.  6z. 

Analysis  :  Since  two  lines  are  ||  if  they  make  equal  exte- 
rior-interior angles  with  any  transversal,  it  is  suggested, 
immediately,  to  draw  any  transversal  to  MN  through  O  and 
at  O  construct  an  angle  equal  to  the  angle  made  by  this 
transversal  with  MN. 

Draw  through  O  the  transversal  AB,  cutting  MN  at  P. 
At  O  on  AB  construct  (with  OA  as  one  side)  an  angle  equal 
to  APN.  Follow  Problem  5. 

Problem  7. 

116.  To  divide  a  straight  line  into  any  number  of  equal 
parts. 

Let  XY  be  the  line;  to  divide  it  into,  say,  5  equal  parts. 

Analysis:  Bearing  in  mind  that  if  any  number  of  parallel 
lines  cut  equal  parts  on  one  transversal  they  cut  off  equal 
parts  on  every  transversal,  if  we  lay  off  5  equal  spaces  on 


Plane  Geometry. 


69 


any  other  line  and,  by  Problem  6,  draw  parallels  from  the 
points  of  division  to  the  given  line,  the  problem  is  solved. 


Fig.  62. 

Proof:  Draw  any  line  XM  from  X,  of  indefinite  length; 
lay  off  5  equal  distances  (any  convenient  distance)  on  XM. 
Join  the  last  point  of  division  N  on  XM  with  the  end  Y, 
of  XY,  and  from  the  other  points  of  division  on  XM  draw 
parallels  to  NY;  they  will  cut  off  equal  distances  on  XY. 
(Why?) 


Problem  8. 

117.  To  find  the  third  angle  0}  a  triangle  when  two  of 
the  angles  are  given. 

Statement:  Let  the  angles  A  and  B  be  two  angles  of  a 
triangle  to  find  the  other  angle. 


Fig.  62a. 


Fig.  6ab. 


Analysis:   The   three   angles   of  every   triangle   together 
make  two  right  angles  or  one  straight  angle.     Then  what 


70  Plane  Geometry. 

remains  of  a  straight  angle  after  the  two  angles  A  and  B 
are  cut  off  from  it  must  be  the  required  angle. 
Complete  the  construction. 

Problem  9. 

118.    To  construct  a  triangle  having  given  two  sides  and 
the  included  angle. 


Fig.  63. 

Statement:  Let  M  and  N  and  angle  A  be  the  given  parts 
to  construct  the  triangle. 

Analysis:  The  angle  will  determine  the  directions  of 
these  two  given  sides,  and  their  extremities  (since  their 
lengths  are  known)  will  determine  the  extremities  of  the 
third  side. 

Proof:  Take  any  indefinite  line  XY  and  lay  off  XZ  =  M 
(or  N).  At  X  construct  the  angle  ZXW  -  Z  A,  and  on 
the  side  XW  lay  off  XV  =  N  (or  M,  if  N  has  been  used 
before).  Join  V  and  Z,  and  XVZ  will  be  the  required 
triangle.  (Why?) 

Problem  10. 

119.  To  construct  a  triangle  having  given  two  angles 
and  the  included  side. 

Statement:  Let  M  and  N  be  the  given  angles  and  O  the 
given  side.  To  construct  the  triangle. 


Plane  Geometry.  71 

Analysis:  The  angles  determine  the  directions  of  the  two 
unknown  sides,  and  since  two  lines  can  intersect  in  only 


one  point,  if  these  lines  have  the  proper  direction,  they 
must  intersect  in  the  required  point. 
Complete  the  construction. 

Problem  11. 

120.  To  construct  a  triangle  having  given  two  sides  and 
the  angle  opposite  one  of  them. 

Statement:  -Let  M,  N,  and  the  angle  O  opposite  N  be 
the  given  parts,  to  construct  the  triangle. 

Analysis:  Since  the  given  angle  must  be  adjacent  to  one 
of  the  given  sides,  it  will  determine  the  direction  of  the 
unknown  side.  The  other  side  being  given  in  length,  and 
starting  at  the  other  end  of  the  given  line  already  used, 
will  cut  this  line  of  unknown  length,  but  known  direction, 
at  the  required  point. 

Proof:  On  the  line  AB  of  indefinite  length  lay  off  AC  = 
M.  At  A  construct  on  AC  the  angle  CAD  =  Z  O. 

With  C  as  a  centre  and  a  radius  equal  to  N,  cut  AD  in 
E  and  F;  then  either  ACF  or  ACE  will  fulfill  the  conditions. 

It  is  plain  that  if  the  arc  described  from  C  with  radius  N 
cuts  AD,  there  will  be  two  triangles  that  fulfill  the  condi- 


72  Plane  Geometry. 

tions;  if  the  arc  just  touches  the  line  AD  there  will  be  only 
one  triangle,  which  will  be  a  right  triangle.  (Why?) 
And  if  the  arc  neither  cuts  nor  touches  the  line  AD,  because 
N  is  less  than  the  distance  from  C  to  AD,  there  is  no  solu- 


Fig.  64. 

tion.     If  it  cuts  AD  only  once  between  A  and  D  there  will 
be  but  one  triangle. 

Problem  12. 

121 .  To  construct  a  triangle  having  three  sides  given. 
Statement:  Let  A,  B  and  C  be  the  three  given  sides,  to 

construct  the  triangle. 

Analysis:  Taking  any  of  the  given  lines  as  a  base, 
arcs  described  from  its  extremities  as  centres  will  intersect 
in  a  point  distant  from  these  extremities  the  lengths  of 
the  other  given  sides. 

Complete  the  construction. 

Problem  13. 

122.  To  construct  a  parallelogram  having  given  two  sides 
and  the  included  angle. 


Plane  Geometry. 


73 


Statement:   Given  the  sides  A  and  B  and  included  angle 
C,  to  construct  the  parallelogram. 


Fig.  65. 

Analysis:  The  opposite  sides  of  a  O  are  parallel,  hence 
if  we  know  the  direction  of  the  two  given  sides  and  their 
lengths,  we  can  readily  draw  the  opposite  sides  (Problem  6). 
Since  we  know  the  angle  between  the  given  sides,  we  know 
their  directions. 

Complete  the  construction. 


Problem   14. 

123.    To  circumscribe  a  circle  about  a  given  triangle. 

Statement:     Given    triangle    MNO,    to   circumscribe 
circle. 

Analysis:  To  circumscribe  a 
circle  about  MNO  it  must  pass 
through  the  vertices  M,  N,  O, 
and  to  secure  that  result  the 
centre  must  be  equally  distant 
from  M,  N  and  O.  The  locus  of  Fig' 66' 

points  equally  distant  from  two  given  points  is  immediately 
suggested. 

Complete  the  construction. 


74  Plane  Geometry. 

Problem  15. 

124.  To  inscribe  a  circle  in  a  given  triangle. 

Analysis:  Here  it  is  a  case  of  finding  a  point  equally 
distant  from  three  given  lines,  which  intersect,  hence  the 
locus  of  points  equally  distant  from  the  sides  of  an  angle  is 
suggested. 

Make  the  construction. 

Problem  16. 

125.  To  draw  a  tangent  to  a  circle  at  a  given  point  (Fig. 
67). 


Analysis:  Since  a  tangent  is  perpendicular  to  a  radius 
at  its  extremity  the  suggestion  is  plain. 

Construction:  Draw  radius  OP  to  point  of  tangency  P; 
a  perpendicular  to  OP  at  P,  its  extremity  (Problem  i, 
Case  II),  AP  is  the  tangent. 


Problem  17. 

126.  To  draw  a  tangent  to  a  circle  from  an  external  point 
(Fig.  68). 

Analysis:  Since  the  point  is  outside  the  circle  we  do  not 
know  where  the  point  of  tangency  will  be,  hence  cannot 


Plane  Geometry.  75 

draw  a  radius  to  that  point,  and  so  cannot  draw  a  perpendi- 
cular as  in  Problem  16.     But  a  right  angle  is  inscribed  in  a 


Fig.  68. 

semi-circle,  and  this  suggests  drawing  a  semi-circle  through 
the  external  point  and  the  centre  of  the  circle,  with  the  line 
from  the  point  to  the  centre  as  diameter. 

Construction:  Join  the  point  P  with  the  centre,  O.  On 
OP  as  diameter  construct  the  semi-circumference  PQO, 
intersecting  the  circle  at  Q.  A  line  joining  P  and  Q  is  tan- 
gent; for  Z_  PQO  being  inscribed  in  a  semi-circle  is  a  right 
angle,  hence  PQ  is  perpendicular  to  OQ  a  radius. 


EXERCISE  III. 

1.  Divide  a  line  12"  long  into  seven  equal  parts. 

2.  Construct  an  angle  of  30°;  of  45°. 

3.  Through  a  given  point  (outside),  to  draw  a  line  mak- 
ing a  given  angle  with  a  given  line. 

4.  Given  base  and  altitude  of  an  isosceles  triangle;  con- 
struct it. 

5.  Given  the  base  and  vertical  angle  of  an  isosceles  tri- 
angle; construct  it. 

6.  Construct  an  equilateral  triangle  having  one  side. 


76  Plane  Geometry. 

7.  Given  the  diagonals  of  a  parallelogram  and  the  angle 
between  them;  construct  the  parallelogram. 

8.  Construct  a  right  triangle  from  its  hypotenuse  and  one 
acute  angle. 

9.  Construct  a  right  triangle  given  hypotenuse  and  one 
leg. 

10.  Construct    an   equilateral    triangle    having   its   peri- 
meter. 

11.  Construct  an  isosceles  triangle  having  its  base  and 
the  sum  of  its  two  equal  sides. 

12.  To  draw  a  tangent  to  given  circle  and  parallel  to 
a  given  line. 

13.  Given  the  altitude  of  an  equilateral  triangle;  con- 
struct the  triangle. 

14.  Through  a  given  point  in  a  circle  draw  the  shortest 
possible  chord. 

15.  Given  an  angle,   an  adjacent  side   (the  base)    and 
the  altitude  of  a  triangle;  construct  it.      Suggestion:  Draw 
a  line  ||  to  the  base  at  the  altitude  distance. 

1 6.  Construct  a  rhombus,  given  base  and  altitude. 

17.  Given  the  diagonal  of  a  square;  construct  the  square. 

18.  Given  the  hypotenuse  of  a  right  triangle  and  the 
length  of  the  perpendicular  upon  it  from  the  right  angle; 
construct  the  triangle. 

19.  Describe  a  circle  of  given  radius  tangent   to   two 
intersecting  lines. 

20.  In  a  given  straight  line  find  a  point  equally  distant 
from  two  intersecting  lines. 

21.  Describe  a   circle  with  a  given  centre  and  tangent 
to  a  given  line. 

22.  Through  a  given  point  outside  a  given  line  draw  a 
circle  tangent  to  the  line  at  a  given  point. 


Plane  Geometry.  77 

23.  Through  a  given  point  inside  a  circle,  to  draw  a  chord 
equal  to  a  given  chord. 

24.  Describe  a  circle  of  given  radius  tangent  to  two  given 
circles. 

25.  Describe  a  circle  touching  two  intersecting  lines,  one 
of  them  at  a  given  point. 

26.  Draw    tangents    common    to    two    non-intersecting 
circles. 

27.  Construct  a  triangle  when  the  mid-points  of  its  sides 
are  given. 

28.  Connect  two  intersecting  lines  by  a  line  that  will  be 
||  to  one  given  line  and  equal  to  another. 

29.  Through  a  given  point  outside  a  given  circle  draw  a 
secant  whose  internal  and  external  segments  will  be  equal. 

30.  Through  a  point  outside  a  circle,  to  draw  a  secant 
line,  so  that  the  segment  of  it  inside  the  circle  shall  be  equal 
to  a  given  line. 

31.  Draw  a  tangent  to  a  given  circle,  making  a  given 
angle  with  a  given  line. 

32.  Show  that  the  diameter  of  a  circle  inscribed  in  a 
right  triangle  equals  the  sum  of  the  two  legs  minus  the 
hypotenuse. 

33.  Construct  a  right  triangle  whose  hypotenuse  is  three 
times  the  length  of  the  shorter  leg. 

34.  Construct  an  isosceles  triangle,  given  base  and  peri- 
meter. 

35.  Given,  in  a  right  triangle,  an  acute  angle  and  the 
sum  of  the  legs;    construct  it. 

36.  Given  hypotenuse  and  the  sum  of  the  legs. 

37.  Construct     a    triangle,    given     altitude    and    base 
angles. 

38.  Construct  a  triangle,  given  the  angles  and  the  sum 
of  two  sides. 


78  Plane  Geometry. 

39.  Given  two  circles   that  intersect;  draw  a  common 
tangent. 

40.  Through  one  of  the  intersection  points  of  two  circles, 
to  draw  a  straight  line,  so  that  equal  parts  of  the  line  will 
lie  inside  of  each  circle. 

41.  Show  by  construction  how  much  f  exceeds  }. 

42.  Extract  the  square  root  of  5  geometrically. 


APPLICATIONS    OF    PROPORTION. 
Proposition    I. 

127.  //  a  line  is  drawn  parallel  to  one  side  of  a  triangle, 
it  divides  the  other  two  sides  proportionally. 

Statement :  If  in  the  triangle 
xyz,  AB  be  drawn  parallel  to  yz, 
then  xA:  Ay  :  :  xE  :  Hz. 

Analysis:  Whenever  a  divi- 
sion into  parts  of  two  geomet- 
rical magnitudes  is  involved, 
there  are  two  cases;  ist,  where 
the  magnitudes  have  a  common  measure;  2d,  where  they 
have  not. 

The  first  case  requires  only  a  comparison  of  the  number  of 
times  the  magnitudes  contain  their  common  unit  of  measure; 
the  second  case  requires  the  use  of  a  common  measure  so 
small  that  the  magnitudes  approach  commensurability. 

Proof  of  ist  case:  Find  the  common  measure  of  JcA  and 
Ay,  and  say  xA  contains  it  m  times  and  Ay,  n  times;  then  ocA : 
Ay  :  :  m  :  n. 

Through  the  points  of  division  draw  lines  parallel  to  yz; 
then  these  lines  divide  xE  into  m  equal  parts  and  Bz  into  n 
equal  parts.  (Why?) 


Plane  Geometry.  yp 

Hence  xA  :  Ay  :  :  m  :  n  and  xE  :  Bz  :  :  m  :  n. 

.•.xA  :Ay  :  :  xB  :  Bz.     Prove  the  second  case. 
128.   Cor.  I.     Taking  the  above  proportion  by  composi- 
tion, 

xA  +  Ay  :  xA  (or  Ay)   :  :  xE  +  Bz  :  xE  (or  Bz). 

Or  xy  :  xA  (or  Ay}  :  :  xz  :  xE  (or  Bz). 
That  is,  a  line  II  to  one 

A  E 

side  of   a  triangle  divides                    /               \ 
the  other  sides  into  parts  / \L 

•pA  V 

which  have  the  same  ratio  /  \  \ 

to  those  sides,  respectively. 


129.    Cor.   II.     If    two 


7    \  \ 

/  \N V 


lines  are  cut  by  any  num-         /  \  \ 

ber  of  parallels,  their  cor-        / \Q \< 

responding   segments    are 

proportional.      That  is,  if 

the  lines  AB  and  ED  are  B 

cut  by  the  parallels,    FL, 

GX,  HY,  KZ,  etc.,  then  Fig'  7°' 

FG  :  LX  :  :  GH  :  XY  :  :  HK  :  YZ,  etc. 

From  F,  where  FL  cuts  AB,  draw  FC    ||   to  ED,  cutting 
GX  at  M,  HY  at  N,  and  KZ  at  O. 
By  Cor.  I,  FG  :  FM  :  :  FH  :  FN  (i) 

GH  :MN:  :  FH  :  FN  (2) 
FH  :FN  :  :FK  :  FO  (3) 

and  HK  :  NO  :  :  FK  :  FO  (4) 

From  (i)  and  (2)  FG  :  FM  :  :  GH  :  MN. 

From  (2),  (3)  and  (4)  GH  :  MN  :  :  HK  :  NO. 

/.  FG  :  FM  :  :  GH  :  MN  :  :  HK  :  NO 
but      FM  =  LX,  MN  =  XY  and  NO  =  YZ.     (Why?) 
/.  FG  :  LX  :  :  GH  :  XY  :  :  HK  :  YZ. 


So  Plane  Geometry. 

Proposition   II. 

130.  //  a  straight  line  divides  two  sides  of  a  triangle  pro- 
portionally, it  is  parallel  to  the  third  side. 

Statement:  If  AB  divides  the  sides  MN  and  MO  of  the 
triangle  MNO,  so  that  MN  :  MB  :  :  MO  :  MA,  it  is  ||  to 
NO. 

Analysis:   If  a  line  is  drawn  parallel  to  NO  through   A 

(say),  we  know  that  it  will 
divide  the  sides  proportion- 
ally, hence  it  is  only  necessary 
to  identify  AB  with  this  line. 

Proof:  Through  A  draw  a 
line  ||  to  NO,  cutting  MN  at 
C  (say).  Then  MN  :  MC  :  : 
MO  :  MA;  but  by  hypothesis 
MN  :  MB  :  :  MO  :  MA. 

Fi8'  7I-  /.  MC  =  MB;  that  is  B  and 

C  are  the  same  point,  and  since  AB  and  AC  have  the  points 
A  and  B  in  common,  they  are  the  same  straight  line. 

Definition. 

131.  If  a  line  AB  be  divided  at  C  between  A  and  B,  it  is 
said  to  be  divided  internally  into  the  segments  AC  and  CB. 
If  C  is  not  between  A  and  B,  but  on  the  prolongation  of  AB, 


Fig.  71  a. 


AB  is  said  to  be  divided  externally  by  C'  into  the  segments 
AC'  and  C'B. 

In  both  cases  the  segments  are  the  distances  from  the 
extremities  of  the  line  AB  to  the  point  C  or  C'   (Fig.  710). 


Plane  Geometry. 


81 


If  a  line  is  divided  internally  and  externally,  so  that  the 
segments  have  the  same  ratio,  the  line  is  said  to  be  divided 
harmonically. 

Proposition   III. 

132.  The  bisector  of  an  angle  of  a  triangle  divides  the 
opposite   side  into  segments   proportional  to  the  including 
sides. 

Statement:  Let  AC  bisect  the  angle  A  of  the  triangle 
ABD,  then  BC  :  CD  :  :  BA  :  AD. 

Analysis:  We  know  that  a  line  parallel  to  one  side  of  a 
triangle  divides  the  other  sides  proportionally,  and  since 
AC  is  the  dividing  line  in 
this  case,  a  triangle  must  be 
formed  by  drawing  a  line 
||  to  AC,  in  order  to  use 
the  above  principle. 

Proof:  From  D  draw  DE 
||  to  AC  intersecting  AB 
produced  at  E. 

By  Prop.  I,  BC  :  CD  :  : 
BA  :  AE  (i). 

But  since  DE  is  ||  to 
AC,  Z  BAG  -  ZAED 
and  Z  CAD  =  Z  ADE  (Why?),  and  Z  BAG  =  Z  CAD. 
.*.  Z  ADE  -  Z  AED,  and  hence  AE  =  AD.  Substituting 
in  (i)  AD  for  AE; 

BC  :  CD  :  :  BA  :  AD. 

Proposition    IV. 

133.  The  bisector  of  the  exterior  angle  of  a  triangle  divides 
the  opposite   side  externally  into  segments  proportional  to 
the  sides  forming  this  exterior  angle. 


82  Plane  Geometry. 

Statement:  If  AB  bisects  the  exterior  angle  CAF  of  tri- 
angle CAD,  intersecting  CD  produced  at  B,  then  BC  :  BD  :  : 
AC  :  AD. 

Analysis:  Same  as  for  last  proposition. 


Proof:  From  C  draw  CE  ||  to  AB,  cutting  AD  at  E;  then 
by  Prop.  I,  BC  :  BD  :  :  AE  :  AD  (i). 

In  the  triangle  AEC  (since  CE  is  ||  to  AB),  Z  ACE  = 
Z  BAG  and  Z  AEC  =  Z  BAF,  but  Z  BAG  -  Z  BAF; 
hence  Z  ACE  =  Z  AEC,  and  .-.  AE  =  AC. 

Substituting  AC  for  AE  in    (i) 

BC  :BD  :  :  AC  :  AD. 

When  is  above  proposition  not  true  ? 

134.  Cor.     The    bisectors    of    the    interior  and  exterior 
angles  at  the  same  vertex  divide  the  opposite   side   har- 
monically. 

Similar  Polygons. 

135.  Dej.I.    Similar  polygons  are  those  which  have  equal 
angles   and    their    homologous    sides    proportional.     Thus 
ABCDE  and  MNOPQ  are  similar  if  Z  A  -  ZM,  ZB  = 
Z  N,  Z  C  =  Z  O,  etc.,  and  AB  :  MN  :  :  AE  :  MQ  :  :  ED  : 
QP,  etc.     This  definition,  of  course,  includes  triangles. 


Plane  Geometry. 


Fig.  74. 

136.  Def.  II.     Homologous  parts  of  similar  polygons  are 
those  similarly  situated.     In  triangles,  for  instance,  homo- 
logous sides  are  those  opposite  the  equal  angles. 

Remark:  It  should  be  noted  that  polygons  that  differ 
greatly  in  size  may  be  similar. 

Similarity  in  the  geometric  sense  applies  only  to  form 
not  to  dimension. 

Proposition  V. 

137.  Two  mutually  equiangular  triangles  are  similar. 
Statement:  If  MNO  and  ABC  are  mutually  equiangular, 


M- 


thatis,  if  Z  M  =  Z  B,  Z  N  =  Z  A,  and  Z  O  =  Z  C,  then 
MNO  and  ABC  are  similar. 

Analysis:  Since  the  triangles  are  equiangular,  it  remains 
only  to  prove  that  their  homologous  sides  are  proportional. 
Since  the  angles  are  equal,  the  sides,  including  any  one  angle, 


84  Plane  Geometry. 

will  take  the  same  direction  if  the  triangles  are  superposed, 
and  then  if  the  remaining  sides  can  be  shown  to  be  ||  ,  Prop. 
I  applies. 

Proof:  Apply  MNO  to  ABC  so  that  the  vertex  N  falls  on 
A;  then  if  the  side  MN  is  made  to  fall  on  AB,  the  side  NO 
will  fall  on  AC  (Why?),  and  the  side  MO  will  take  the 
position  indicated  in  the  figure,  joining  the  points  M  and  O, 
lying  respectively  on  AB  and  AC. 

Since  all  the  angles  of  MNO  are  equal  to  the  corre- 
sponding angles  of  ABC,  Z  AOM  (corresponding  to  Z  O) 

=  ZC. 

.'.  MO  is  ||  to  BC.  (Why?) 

Hence  AM  :  AB  :  :  AO  :  AC.  (Why?) 

But  AM  =  NM  and  AO  =  NO. 

.-.  NM  :  AB  :  :  NO  :  AC. 

If  the  Z  O  is  made  to  coincide  with  Z  C  it  can  be  shown 
in  the  same  way  that  MO  :  BC  :  :  NO  :  AC. 

138.  Cor.  I.     Two  triangles  are  similar  if  two  angles  of 
the  one  are  equal  to  two  angles  of  the  other.          (Why?) 

139.  Cor.  II.     Two  right  triangles  are  similar  if  an  acute 
angle  of  one  is  equal  to  an  acute  angle  of  the  other. 

Proposition    VI. 

140.  //  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other,  and  the  including  sides  proportional,  they 
are  similar. 

Sta'ement:  If  in  the  triangles  xyz  and  ABC,  Z#  =  /-A 
and  xy  :  AB  :  :  xz  :  AC,  xyz  and  ABC  are  similar. 

Analysis:  Since  one  angle  in  each  is  equal,  the  including 
sides  will  take  the  same  directions,  when  the  triangles 
superposed,  the  third  side  of  the  smaller  cutting  the  two 
sides  including  the  equal  angle  of  the  other,  proportionally 


Plane  Geometry.  85 

by  hypothesis,  for  the  including  sides  in  the  smaller  tri- 
angle will  form  segments  of  the  corresponding  sides  of  the 
larger. 

Proof:  Apply  xyz  to  ABC,  making  angle  x  coincide  with 
ZA,  and  hence  the  side  xz  falls  on  AC  and  xy  on  AB. 


ZT"  \ 


Fig.  76. 

Then,  since  xy  :  AB  :  :  xz  :  AC, 

Ay  :  AB  :  :  A2  :  AC. 

Hence  yz  is  parallel  to  BC  (Why  ?)  ;  but  if  yz  is  ||  to  BC 
Z  Ayz  =  Z  B  and  Z  Azy  =  Z  C,  that  is,  Zy  =  Z£  and 
Zz  =  Z  C,  since  Ayz  is  simply  the  triangle  xyz  placed 
on  ABC. 

.'.  xyz  and  ABC  are  equiangular;  they  are  similar  (Prop.V). 

Proposition    VII. 

141.    Two  triangles  whose  sides  are  respectively  propor- 
tional are  similar. 
2 


Fig.  77- 

Statement:  If  in  the  triangles  123  and  456,  12:  45:  123  : 
56  :  :  13  : 46,   the   triangles   are   similar. 


86 


Plane  Geometry. 


Analysis:  Since  no  angles  are  given  equal,  we  cannot 
directly  apply  with  effect  one  triangle  to  the  other,  but  we 
can  lay  off  the  sides  of  the  smaller  upon  those  of  the  larger, 
and  connecting  the  points  thus  determined,  attempt  to 
identify  this  triangle  so  constructed  with  the  given  triangle. 

Proof:  On  12  lay  off  27  —  54,  and  on  23  lay  off  28  =  56. 
Join  7  and  8,  forming  the  triangle  278. 

The  triangle  278  is  similar  to  123.  (Why?) 

Also  in  the  triangles  278  and  123  (since  they  are  similar) 
21  :  27  :  :  13  :  78.  (a) 

But  by  hypothesis,  21:  45  :  :  13  :  46,  (b) 

27  =  45  by  construction,  hence,  by  comparing  (a)  and  (b) 
78  =  46. 

.'.  triangle  278  and  triangle  456  (having  3  sides  equal)  are 
equal  triangles.  Hence,  since  278  is  similar  to  123,  456  is 
similar  to  123. 

Proposition  VIII. 

142.  Two  triangles  having  their  sides  respectively  parallel 
or  perpendicular  are  similar. 

B 


Fig.  78. 

Statement:   If  in  the  triangles  ABC  and  DEF,  AB  is  ||  to 

ED,  AC  ||  to  DF,  andBC  11  to  EF,  the  triangles  are  similar. 

Also,  in  the  triangles  MNO  and  XYZ,  if  XY  is  perpendi- 


Plane  Geometry.  87 

cular  to  MO,  YZ  perpendicular  to  MN,  and  XZ  perpen- 
dicular to  NO,  MNO  and  XYZ  are  similar. 

Analysis:  Since  angles  whose  sides  are  either  ||  or  per- 
pendicular are  equal  (provided  the  sides  extend  in  the  same 
direction  as  these  must  do)  (Why?),  these  triangles  will  be 
equiangular  and  hence  similar. 


Proposition   IX. 

143.  Two  homologous  altitudes  of  two  similar  triangles 
have  the  same  ratio  as  any  two  homologous  sides. 

Statement:  If  ABC  and  MNP  are  similar,  and  BD  and  NO 
are  the  altitudes,  then  BD  :  NO  :  :  AB  :  MN  :  :  AC  :  MP  :  : 
BC  :  NP. 


Fig.  79- 

Analysis:  The  altitudes  form  two  right  triangles  which 
are  readily  proved  similar. 

Proof:   ABD  and  MNO  are  similar.  (Why?) 

.-.  AB  :  MN  :  :  BD  :  NO  (i).  Also  BDC  and  NOP  are 
similar. 

/.BC  :NP  :  :  BD  :  NO  (2). 

Also  AB  :MN::AC:MP   (3). 

Then  BD  :  NO  :  :  AB  :  MN  :  :  AC  :  MP  :  :  BC  :  NP, 
[combining  (i),  (2)  and  (3)]. 


88 


Plane  Geometry. 


144.   Cor.  Any  homologous  lines  in  two  similar  triangles 
are  in  proportion  to  the  sides. 


Proposition  X. 

//  two  parallels  are  cut  by  three  or  more  tranversals  that 
pass  through  the  same  point,  the  corresponding  segments 
are  proportional. 

Statement:  Let  A  be  the  fixed  point  (whether  between 
the  parallels  or  on  the  same  side  of  both),  and  BF  and  GM 
be  the  parallels  cutting  the  transversals,  at  B,  C,  D,  E  and 
F,  and  G,  H,  K,  L  and  M  respectively. 

Analysis:  The  similar  triangles  are  evident. 

Proof:  Since  ABC  and  AGH  are  similar  (Why?),  BC  : 
GH  ::  AC  :  AH.  Also  since  ACD  and  AHK  are  similar, 
AC:  AH::  CD  :  HK.  .'.  BC  :  GH  ::  CD  :  HK. 


G        H     K     L       M 


Fig.  80. 


In  the  same  way  may  be  proved  that  CD  :  HK  ::  DE  : 
KL,  etc. 

Hence  BC  :  GH  ::  CD  :  HK  ::  DE  :  KL  ::  EF  :  LM. 


Plane  Geometry. 
Proposition  XI. 


Fig.  81. 


145.  //  three  or  more 
non-parallel  transversals 
intercept  proportional 
segments  upon  two  par- 
allels,  they  pass  through 
a  common  point. 

Statement:  If  BE,  CF 
and  DG  cut  MN  and 
RS  at  B,  C,  D  and  E, 
F,  G  respectively  so  that 
BC  :  EF  ::  CD  :  FG, 
then  BE,  CF,  DG  pro- 
duced, will  meet  in  a 
point,  say  A. 

Analysis:  Two  non-parallel  lines  will  always  meet  if 
produced,  and  if  a  third  line  be  drawn  through  this  point 
and  one  of  the  intersection  points  of  a  third  given  trans- 
versal, this  third  given  transversal  may  be  identified  with 
the  arbitrarily  drawn  third  line,  and  hence  it  will  pass 
through  A. 

Proof:  Produce  BE  and  CF  to  meet  in,  say,  A.  Through 
A  and  G  draw  arbitrarily  a  line,  which  meets  MN,  say,  at 
O.  Then  by  Prop.  IX, 

BC  :  EF  ::  CO  :  FG, 

but  by  hypothesis,  BC  :  EF  ::  CD  :  FG. 

/.  CO  =  CD,  that  is,  the  points  O  and  D  are  the  same; 
hence  the  arbitrary  line  AOG  is  the  same  as  the  given  line 
DG. 

.*.  DG  will  pass  through  A. 


9o 


Plane  Geometry. 


Proposition  XII. 

146.    The  perimeter   of  two   similar   polygons   have  the 
same  ratio  as  any  two  homologous  sides. 


Fig.   8a. 

Statement:  If  ABODE  and  MNOPQ  are  similar,  then 
AB  +  BC  +  CD  +  DE  +  AE  :  MN  +  NO  +  OP,  etc., 
::  AB  :  MN  ::  AE  :  MQ,  etc. 

Analysis:  The  sides  form  a  series  of  equal  ratios;  hence 
by  the  theory  of  proportion,  the  antecedents  and  conse- 
quents may  be  combined. 

Proof:   Since  the  polygons  are  similar  by  definition, 
AB  :  MN  ::  AE  :  MQ  ::  ED  :  QP,  etc. 

.'.  (AB  +  AE  +  ED  +  etc.)  :  (MN  +  MQ  +  QP  + 
etc.)  ::  AB  :  MN  ::  AE  :  MQ,  etc.,  by  theory  of  proportion. 


Proposition  XIII. 

147.  //  in  a  right  triangle  a  perpendicular  is  drawn  from 
the  vertex  of  the  right  angle  to  the  hypotenuse: 

The  triangles  thus  formed  are  similar  to  each  other  and 
to  the  given  triangle. 

The  perpendicular  is  the  mean  proportional  between  the 
segments  of  the  hypotenuse. 


Plane  Geometry. 


91 


G 


Fig.  83. 


Each  leg  of  the  right  triangle  is  the  mean  proportional 
between  the  hypotenuse  and  the  adjacent  segment. 

Statement :  Let  AC  be  a  JL  from 
the  right  angle  A  upon  the  hy-  A 

potenuse  BD,  then  triangles  BAG 
and  ACD  are  similar  to  each 
other  and  to  the  triangle  BAD; 
also  BC  :  AC  ::  AC  :  CD.B, 
BD  :AB  ::  AB  :  BC  and  BD  : 
AD  ::  AD  :  CD. 

Analysis:  To  prove  two  triangles  similar,  it  is  usually 
easiest  to  show  that  they  have  equal  angles.  After  they 
are  proved  similar,  their  sides  are  known  to  be  proportional. 

Proof:  In  triangle  ABC  and  triangle  ACD,  Z  BAG  - 
Z  ADC  (give  two  reasons).  .'.  since  they  are  right  triangles, 
triangle  ABC  is  similar  to  triangle  ACD.  Triangle  ABC 
and  triangle  ABD  have  Z_  B  common,  .'.  they  are  similar, 
and  triangle  ACD  and  ABD  have  Z  D  common,  .'.  they 
are  similar. 

In  the  similar  triangles  ABC  and  ACD,  BC  :  AC  ::  AC 
:  CD.  Also  in  the  similar  triangles  ABC  and  ABD  (see 
note)  BD  :  AB  ::  AB  :  BC.  In  the  similar  triangles  ACD 
and  ABD,  BD  :  AD  ::  AD  :  CD. 

NOTE.  —  To  make  the  relations  clear  this  proportion 
may  be  written  thus:  BC  (opposite  x  in  ABC)  :  AC 
(opposite  x  in  ACD)  ::  AC  (opposite  y  in  ABC)  :  CD 
(opposite  y  in  ACD).  Likewise,  BD  (opposite  z  in  BAD) 
:  AB  (opposite  z  in  ABC)-::  AB  (opposite  x  in  ABD)  :  BC 
(opposite  x  in  ABC). 

It  will  be  observed  that  homologous  sides  of  similar  tri- 
angles are  those  opposite  equal  angles,  hence  it  facilitates 
the  selection  of  homologous  sides  to  mark  the  equal  angles 
with  the  same  letter,  as  x,  y  or  z. 


92  Plane  Geometry. 

148.  Cor.  I.    The  perpendicular  from  any  point  in  the 
circumference  of  a  circle  is  a  mean  proportional   between 
the   segments   of   the   diameter. 

149.  Cor.  II.   The  square  of  the  hypotenuse  is  equal  to 
the  sum  of  the  squares  of  the  other  two  sides. 

For  BD  :  AB  ::  AB  :  BC  or  BD  X  BC  =  AB2 
and  BD  :  AD  ::  AD  :  CD  or  BD  X  CD  ="AD2 
add  BD  (BC  +  CD)  =  AB2  +  AT)2. 

But  BC  +  CD  =  BD. 

v  BD2  =  AB2  +  AD2. 

150.  Definition.   The  projection  of  a  point  on  a  line  is  the 
foot  of  the  perpendicular  from  the  point  to  the  line.     The 
projection  of  one  line  on  another  is  the  segment  of  the  second 
line  included  between  the  feet  of  the  perpendiculars  upon  it 
from  the  extremities  of  the  first  line. 

If  the  lines  intersect,  the  projection  of  either  line  upon  the 
other  is  the  distance  from  the  point  of  intersection  to  the 
foot  of  the  perpendicular,  from  the  end  of  the  projected  line 
upon  the  other. 


Proposition  XIV. 

151.  In  any  triangle  the  square  oj  the  side  opposite  an 
acute  angle  is  equal  to  the  sum  oj  the  squares  oj  the  other 
two  sides  diminished  by  twice  the  product  of  one  of  those 
sides  by  the  projection  oj  the  other  upon  that  side. 

Statement:  In  the  triangle  ABD,  when  CD  is  the  projec- 
tion of  BD  on  AD,  then 

AB2  =  AD2  +  BD2  -  2  AD  X  CD. 


Plane  Geometry. 


93 


Analysis:  In  cases  of  this  sort  it  is  the  most  direct  process 
to  find  the  direct  value  of  the  line  involved  (here,  AB2)  by 
the  most  evident  process  and  then  to  reduce  this  value  to  the 
required  form. 


Proof:  AB2  =  AC2  +  BC2. 
Cor.  II.) 

But   AC  =   AD  - 

CD.  (This  expres- 
sion is  taken  to  bring 
in  the  whole  line  AD 
and  the  other  segment 
CD,  both  of  which  are 
given  in  the  required 
expression.) 


(a)  (By  Prop.  XIII, 


Fig.  84. 


•'•      1C2  =  AD2  +  Cff  ~  *  AD  X  CD. 
Substituting  this  value  of  AC2  in  (a),  AB2  =  AD2  +  CD2 
+  B€2  -  2  AD  X  CD.    But  CD2  +  BC2  =  BD2  (Why?). 
.-.  AB2  =  AD2  +  BD2  -  2AD  X  CD. 

Cor.  If  the  triangle  contains  an  obtuse  angle,  then 
the  square  of  the  side  opposite  the  obtuse  angle  equals 
the  sum  of  the  squares  of  the  other  two  sides  plus  the 
product  of  one  of  those  sides  by  the  projection  of  the 
other  upon  it. 

The  proof  is  exactly  similar,  except  that  here  one  seg- 
ment of  the  side  upon  which  projection  is  made  is  equal 
to  the  whole  side  plus  other  segment. 


94 


Plane  Geometry. 


Proposition  XV. 

152.  The  sum  of  the  squares  of  two  sides  0}  a  triangle  is 
equal  to  twice  the  square  of  half  the  third  side  plus  twice 
the  square  of  the  median  upon  that  side. 

The  difference  of  the  squares  of  two  sides  of  a  triangle  is 
equal  to  twice  the  product  of  the  third  side  by  the  projection 
of  the  median  upon  it. 

Statement:  In  the  triangle  ABE, 
let  BC  represent  the  median  and 
BD  be  a  -L  from  B  upon  AE,  then 
CD  is  the  projection  of  BC  upon 
AE. 
Then 

AB2  +  BE2  =2EC*  +2  AC2  and  AB2  -  BE2=2  AEXCD. 

Analysis:  Same  as  last  proposition. 
Proof: 

BE2  =  BC2  +  CE2  -  2  CE  X  CD  (Prop  XIV). 
AB2  =  BC2  +  AC2  +  2  AC  X  CD  (Prop.  XIV,  Cor.) 
Add,  remembering  that  CE  =  AC, 

BE2  +  AB2  =  2  BC2  +  2  AC2/ 

Subtract  ; 

AB2-BE2=2  CE  X  CD  +  2  AC  X  CD  =  CD  X  4  AC= 

CD  X  2  AE. 

The  median,  which  is  a  line  drawn  from  one  vertex  in 
a  triangle  to  the  middle  of  the  opposite  side,  should  be 
carefully  distinguished  from  the  bisector  of  an  angle,  which 
does  not  usually  meet  the  middle  of  the  side  opposite  to 
the  angle. 


Plane  Geometry.  g«j 

Proposition  XVI. 

153.  //  two  chords  intersect  in  a  circle,  the  product  of  the 
segments  of  one  is  equal  to  the  product  of  the  segments  of 
the  other. 

Statement:  Let  the  chords  MN  ^^ 

and  PQ  in  the  circle  C,  intersect 
at  O,  then  to  prove  MO  X  ON  = 
OP  X  OQ. 

Analysis:  The  required  equa- 
tion, MO  X  ON  =  OP  X  OQ, 
suggests  the  product  of  means 
and  extremes,  and  hence  a  pro-  **&•  86< 

portion,    and    a    proportion    suggests   similar  triangles. 

Then  it  is  necessary  to  form  similar  triangles,  which  con- 
tain the  lines  MO,  ON,  OP,  and  OQ  as  sides.  The  aux- 
iliary lines  required  are  evident. 

Proof:  Connect  M  with  P,  and  N  with  Q,  then  in  the 
two  triangles  MOP  and  NOQ: 

Z  MOP  =  Z  NOQ.  (Why?) 

Z  PMO  =  Z  NQO.  (Why?) 

Z  MPO  =  Z  ONQ.  (Why?) 

.'.  MOP  is  similar  to  NOQ.  Mark  equal  angles  as 
*x,  y  and  z  in  both. 

Then  PO  (opposite  x  in  MOP) :  ON  (opposite  x  in  NOQ) 
:  :  MO  (opposite  y  in  MOP)  :  OQ  (opposite  y  in  NOQ). 

PO  :  ON  :  :  MO  :  OQ. 

/.  PO  X  OQ  =  ON  X  MO  (equating  products  of  ex- 
tremes and  means). 


Fig.  87. 


Plane  Geometry. 

Proposition  XVII. 

154.  //  from  a  point  without  a 
circle,  a  secant  and  a  tangent  are 
drawn,  the  tangent  is  the  mean  pro- 
portional between  the  whole  secant 
and  the  part  of  it  without  the  circle 
(its  external  segment). 

Statement:  Let  A  be  the  point,  AD 
the  secant,  and  AB  the  tangent  to 
circle  C. 

Then 

AD  :  AB  :  :  AB  :  AE.  (AE  being 
part  of  secant  outside  circle.) 

Analysis:  Same  as  for  last  propo- 
sition. 


Proof:  Join  B  with  E,  and  B  with  D. 
Then  in  the  A's  ABE  and  ABD. 


ZAEB 


Z  A  =  Z  A. 

Z  ABE  =  Z  ADB.  (Why?) 

Z  ABD  (Why?),  marking  x,  y  and  z. 


Then  AD  (opp.  z  in  ABD)  :  AB  (opp.  2  in  ABE)  :  :  AB 
(opp.  y  in  ABD)  :  AE  (opp.  y  in  ABE)  or  AD  X  AE  =  AB2. 

Remark:  The  length  of  a  secant  is  understood  to  mean  the 
distance  from  the  external  point  to  the  second  point  of 
intersection  with  the  circle. 


EXERCISE  IV. 

i.  Show  that  in  a  given  circle  the  product  of  the  parts 
into  which  a  fixed  point  divides  any  chord  drawrn  through 
it,  is  constant. 


Plane  Geometry.  97 

2.  If  the  sides  of   a  quadrilateral  are  bisected    prove 
that  the  lines  joining  these  middle  points  in  order  form  a 
parallelogram. 

Suggestion:  Draw  the  diagonals. 

3.  With  a  pole  ten  feet  long,  show  how  to  find  the  height 
of  a  steeple  on  a  clear  day. 

4.  With  sufficient  level  space  on  the  bank  of  a  stream 
show  how  to  find  its  width,  by  using  a  tape  line  only. 

5.  Prove  that  the  lines  joining  the  feet  of  -L's  from  the 
vertices  of  a  parallelogram  upon  the  opposite  diagonals, 
themselves  form  a  parallelogram. 

6.  Show  that  the  medians  of  a  triangle  meet  in  a  point. 

7.  The  common  external  tangent  of  two  circles,  touch- 
ing externally,  is  a  mean  between  their  diameters. 

8.  If  a  quadrilateral  is  inscribed  in  a  circle,  the  product 
of  its  diagonals  equals  the  sum  of  the  products  of  opposite 
sides. 

9.  If  three  lines  be  drawn  through  the  point  of  tangency 
of  two  circles,  the  triangle  formed  by  joining  their  other 
points  of  intersection  in  each  circle  is  similar. 

10.  In  the  parallelogram  mnop,  pr  is  drawn,  meeting  the 
side  mn,  produced  at  r  ;  the  diagonal  om  at  s  and  on  at  t. 

Prove:  ps2  =  st  X  sr. 

n.   Tangents  drawn  to  two  intersecting  circles  forming 
a  point  on  their  common  chord  produced,  are  equal. 

12.  Two   circles  intersect.     In   each   circle   a  chord  is 
drawn  from  one  of  the  points  of  intersection,  tangent  to 
the  other  circle.     Show  that  the  common  chord  is  a  mean 
proportional  between  the  lines  joining  the  other  point  of 
intersection  with  the  ends  of  the  tangent  chords. 

13.  Determine  a  point  in  the  circumference  of  a  circle 
from  which  if  chords  be  drawn  to  two  fixed  points,  these 
chords  will  have  the  ratio  2:3. 


V, 

98  Plane  Geometry. 

14.  In  a  given  circle  a  line  is  drawn  perpendicular  to 
the  diameter  produced.     If  from  the  remoter  end  of  the 
diameter,  a  secant  be  drawn  to  any  point  of  the  perpen- 
dicular,  the  product  of  the  whole  secant  by  its  internal 
segment  is  constant. 

15.  To  pass  a  circle  through  two  given  points  tangent 
to  a  given  line. 

16.  If  there  be  two  luminous  points,  whose  powers  of 
illuminating  a  point  equally  distant  from  each  are  as  3  14, 
find  the  locus   of    points   in   their    plane    equally    illumi- 
nated. 

17.  If  the  middle  points  of  the  two  pairs  of  adjacent  sides 
of  a  parallelogram  are  joined  by  straight  lines,  show  that 
the  diagonal  is  trisected. 

1 8.  Through  a  given  point  within  a  given  circle  draw  a 
chord   which   shall   be   divided  in   the  ratio  i   :  2  by  the 
point. 

19.  Describe   a   circle   whose   circumference   shall   pass 
through  a  given  point  and  cut  two  given  circles  orthogonally, 
that  is,  at  right  angles. 

20.  If   through    the    point   of    tangency  of   two   circles 
three  lines  are  drawn,  prove  that  the  triangles  formed  by 
joining    the   three   intersection   points  in   each   circle  are 
similar. 

21.  The  sides  of  a  triangle  are  8  ft.,  15  ft.,  and  17  ft.,  the 
shortest  side  of  a  similar  triangle  is  5  ft.,  find  the  remaining 
sides.     What  kind  of  a  triangle  is  it  ? 

22.  Find  the  product  of  the  segments  of  a  chord  drawn 
through  a  point  5  in.  from  the  centre  of  a  circle  whose 
radius  is  9  in. 

23.  The  diameter  which  Insects  a  chord  8  ft.  long,  is  ^ 
ft.  in  length.     What  is  the  distance  from  the  end  of  the 
chord  to  the  end  of  the  diameter? 


Plane  Geometry.  99 

24.  The  distance  of  the  moon  from  the  earth  is  approxi- 
mately 239,000  miles.     If  a  disk  i  in.  in  diameter  placed 
10  ft.  from  the  eye  just  covers  the  moon,  what  is  the  moon's 
diameter. 

25.  If  the  diameter  of  the  earth  is  7,918  miles,  what  is 
the  distance  from  the  eye  of  an  observer,  raised  i  mile  above 
the  surface,  to  the  most  remote  visible  point  on  the  earth's 
surface  ? 

26.  The  horizontal  distance  from  A  to  B  is  12.84  ft.,  and 
the  heights  of  A  and  B  above  a  given  level  are  47 .3 2  ft.  and 
55.91    ft.   respectively.     Find  the   height   of  points  whose 
horizontal  distances  from  A  are  i  ft.  and  2  ft.  respectively. 

27.  Divide  a  hexagon  into  4  equal  and  similar  parts. 

28.  Devise  a  method  of  showing  to  the  eye  that  the  sum 
of  the  squares  upon  the  base  and  vertical  leg  of  a  right  tri- 
angle equals  the  square  on  the  hypotenuse. 

29.  A  strip  of  carpet  i  yard  wide  is  laid  diagonally  across 
a  room  18  ft.  square.     What  is  the  length  of  the  strip,  if  it 
just  touches  the  walls  at  its  corners  ? 

30.  The  sides  of  a  triangle  are  24  in.,  25  in.,  and  18  in. 
Find  the  segments  into  which  the  bisector  of  the  angle, 
between  the  24  in.  and  25  in.  sides,  divides  the  18  in.  side. 

31.  Find  the  lengths  of  the  medians  in  Ex.  30. 

32.  The  middle  point  of  a  chord  is  distant  5  in.  from  the 
middle  of  its  subtended  arc.    Find  the  diameter  of  the  circle 
if  the  chord  is  20  in.  long. 

33.  If  another  chord  in  the  same  circle  is  distant  9  in. 
from  the  middle  of  its  subtended  arc,  what  is  the  length  of 
the  chord? 

34.  The  length  of  one  diagonal  of  a  trapezoid  is  30  in., 
and  the  segments  into  which  it  divides  the  other  diagonal 
are  12  in.  and  6  in.     What  are  the  lengths  of  the  segments 
of  the  first  diagonal  ? 


ioo  Plane  Geometry. 

35.  If  two  of  the  adjacent  sides  of  a  parallelogram  are 
5  in.  and  12  in.,  and  the  diagonal  drawn  from  their  point  of 
intersectipn  is  17  in.,  find  the  other  diagonal. 

36.  The  diameters  of  two  circles  are  12  in.  and  28  in., 
respectively,    and    the   distance   between    their   centres   is 
29    in.     What    is    the  length  of    their    common    internal 
tangent. 

37.  Show  that  the  square  of  the  common  tangent  to  two 
circles    (unequal)    touching   each   other   externally   equals 
four  times  the  product  of  their  radii. 

38.  The  diameter  of  a  circle  is  18  in.     One  segment  of  a 
chord  drawn  through  a  point  7  in.  from  the  centre  is  4  in. 
Find  the  other  segment. 

39.  The  diameter  of  a  circle  is  24  in.;  if  tangents  are 
drawn  from  a  point  20  in.  from  the  centre,  what  is  their 
length  and  what  is  the  length  of  the  chord  of  contact  ? 

40.  A  circle  has  a  radius  of  7  in.     From  the  end  of  a 
tangent  24  in.  long  a  secant  is   drawn  through  the  centre. 
What  is  the  length  of  the  secant? 

41.  A  chord  of  a  circle  is  20  in.  long  and  the  sagitta  of  its 
arc  is  2  in.     What  is  the  radius. 

42.  The  span  of  a  circular  arch  is  120  ft.,  and  the  radius 
of  the  circle  of  which  it  is  an  arc  is  725  ft.     Find  the  height 
of  the  middle  of  the  arch. 

AREAS. 

155.  Definition.  The  area  of  a  surface  is  the  number  of 
times  it  contains  the  unit  of  area. 

The  square  whose  side  is  a  linear  unit  in  length  is  accepted 
as  the  unit  of  area. 

That  is,  if  the  area  of  a  surface  is  to  be  expressed  in 
square  inches,  a  square  whose  side  is  one  inch  is  the  unit. 
If  the  area  of  the  surface  is  to  be  expressed  in  square  miles, 


Plane  Geometry. 


101 


a  square,  whose  side  is  one  mile  is  taken  as  the  unit.  Evi- 
dently a  square  of  any  size  desired  may  be  taken  as  a  unit, 
and  the  area  expressed  in  the  square  of  that  unit. 


Proposition  I. 

156.  The  area  of  a  rectangle  equals  the  product  of  its 
base  by  its  altitude. 

That  is,  the  area  of  a  rectangle  in  square  units  equals 
the  product  of  the  number  of  linear  units,  in  its  base  and 
altitude. 

Statement:  If  the  base  of  the  rectangle  ABDC  is  CD  and 
its  altitude  BD,  then  area  of  ABCD  =  BD  X  CD. 

Analysis:  There  are   A  n 

plainly  two  cases;  when 
the  base  and  altitude  con- 
tain a  common  unit,  that 
is,  are  commensurable;  and 
when  they  do  not,  or  are  in- 
commensurable. In  the 
first  case  it  is  only  neces- 
sary to  divide  the  base  and 
altitude  into  as  many  parts  as  they  will  contain  their  com- 
mon unit,  and  by  drawing  through  these  points  of  division, 
lines  ||  to  the  base  and  altitude  respectively,  to  divide  the 
rectangle  into  squares. 

In  the  2d  case,  a  unit  is  chosen  so  small  that  the 
base  and  altitude  approach  commensurability  as  near  as 
desired. 

Proof:  Apply  the  common  unit  to  BD  and  CD.  Say 
it  is  contained  in  BD,  m  times,  and  in  CD,  n  times. 
Through  these  points  of  division  draw  lines  ||  to  BD  and  CD 
respectively.  The  rectangle  will  then  be  divided  into  n 


Fig.   88. 


IO2  Plane  Geometry. 

tiers,  containing  m  squares  each,  that  is  mn  square  units. 
/.  area  of  ABCD  =  mn  =  BD  X  CD. 

Case  II.  Say  the  unit  of  length  chosen  is  contained  in 
BD,  m  times,  and  in  CD,  n  times  plus  a  remainder  ED, 
less  than  the  unit.  Through  E  draw  EF  ||  to  BD,  then 
ACEF  =  EF  X  CE  or  BD  X  CE  (since  EF  =  DB). 

Since  the  remainder  is  always  less  than  the  divisor,  if  we 
take  a  unit  J  as  large  as  before,  it  will  be  contained  in  BD 
and  CE  an  even  number  of  times  and  will  doubtless  reduce 
the  remainder  ED.  If  a  unit  of  indefinite  smallness  be 
used,  the  remainder  ED  will  be  still  smaller,  and  it  is  evi- 
dent that  this  remainder  may  be  made  as  small  as  desired, 
by  taking  a  unit  small  enough.  But  always,  ACEF  = 
BD  X  CE,  hence,  since  E  can  be  made  to  approach  D  as 
near  as  desired,  eventually  ACDB  -  BD  X  CD. 

157.  Cor.  I.   Two    rectangles     having    one     dimension 
equal,  are  to  each  other  as  the  other  dimension. 

158.  Cor.  II.     Two  rectangles   are  to  each  other  as  the 
product  of  their  two  dimensions. 

Proposition  II. 

159.  The  area  oj  a  parallelogram  equals  the  product  of 

its  base  and  altitude. 

Statement:  If  DC  is  the 
base  and  DE  is  the  alti- 
tude of  the  parallelogram 
ABCD,  then  area  of  ABCD 
=  DC  X  DE. 
Fig-  8g-  Analysis:  If  a  rectangle 

be  constructed  on. DC  as  base  and  with  the  altitude  DE, 
by  producing  AB  and  dropping  a  perpendicular  CF  upon  it 
from  C,  the  area  of  the  parallelogram  may  be  compared  with 
this  rectangle  having  the  same  base  and  altitude. 


A-  E 


Plane  Geometry.  103 

Proof:   Construct  the  rectangle  EDCF  as  indicated.    • 

The  area  of  EDCF  =  DC  X  ED. 

In  the  right  triangles  AED  and  BFC. 

ED  =  FC.  (Why?) 

AD  =  BC.  (Why?) 

/.  triangle  AED  =  triangle  BFC.  (Why?) 

But  the  quadrilateral  ADCF  —  triangle  BFC  =  parallelo- 
gram ABCD; 

and  the  quadrilateral  ADCF  —  triangle  AED  =  rectangle 
EDCF. 

/.  O  ABCD  -  rectangle  EDCF.  (Why?) 

But  O  ABCD  has  same  base  and  altitude  as  rectangle 
EDCF,  and  EDCF  =  DC  X  ED.  .'.  ABCD  =  DC  X 
ED. 

State  Corollaries  like  those  for  Prop.  I. 

Proposition  III. 

160.    The  area  of  a  triangle  equals  one-half  the  product 
of  its   base  and  alti- 
tude. 

Statement:  If  ON 
is  the  base  and  MR 
the  altitude  of  the 
triangle  MNO,  then 
area  of  MNO  =  i 
(ON  X  MR).  Fig-  90> 

Analysis:  Recalling  the  relation  between  a  parallelogram 
and  the  two  triangles  into  which  it  is  divided  by  a  diagonal, 
the  procedure  is  evident. 

Proof:  On  ON  as  base  and  with  MR  as  altitude,  con- 
struct a  parallelogram  OMPN  by  drawing  MP  ||  and  equal 
to  ON,  and  joining  P  and  N.  MN  will  evidently  be  the 
diagonal  of  this  parallelogram. 


104 


Plane  Geometry. 


Triangle    OMN  =  J    parallelogram    OMPN,    but   O 
OMPN  =  ON  X  MR  (product  of  base  and  altitude). 
.'.  triangle  OMN  =  4  (ON  X  MR). 

161.  Cor.  I.  By  the  associative  law  of  multiplication, 
4  (ON  X  MR)  =  J  ON  X  MR  =  ON  X  i  MR;  that  is, 
the  area  of  a  triangle  equals  the  product  of  one-half  the  base 
by  the  altitude,  or  the  product  of  one-half  the  altitude  by 
the  base. 
State  corollaries  similar  to  those  under  Prop.  I. 

162.   Cor.  II.   The   area  of   tri- 
angle equals  one-half  the  perimeter 
by  the  radius  of  the  inscribed  circle. 
Call    p    the    perimeter    of   ABC 
and  r  the  radius,   of  the  inscribed 
circle   O.      Draw    radii,    OD,   OE 
and  OF,  to  the  points  of  contact  of 
this  circle,  with  sides,  AC,  AB  and 
BC  respectively.     Also  join  O  with  A,  B  and  C. 

Then  OD  is  the  altitude  of  the  triangle  AOC 


and 

Hence  area 
area 

and  area 
adding;  area 


OE  is  the  altitude  of  the  triangle  AOB  [-(Why?) 
OF  is  the  altitude  of  the  triangle  BOC 

AOC  =  4  (AC  X  OD)  =  \  AC  X  OD  ; 

AOB  =  4  (AB  X  OE), 

BOC  =  4  (BC  X  OF) 


ABC  =  4  (AC  +  AB  +  BC)  X  OD    (since 
OD  =  OE  =  OF  =  r) 


or  area  ABC 


Proposition  IV. 

163.    The  area  of  a  trapezoid  equals  the  product  of  its 
altitude  by  one-half  the  sum  of  its  parallel  sides. 

Statement:   If  AB  and  CD  are  the  two  parallel  sides  of 


Plane  Geometry.  105 

the  trapezoid  ABCD,  and  BE  its  altitude,  then  area  of 
ABDC  =  i  (AB  +  CD)  X  BE. 

Analysis:  A  diagonal 
will  clearly  divide  the 
trapezoid  into  two  trian- 
gles, both  having  the  alti- 
tudes of  the  trapezoid, 
and  the  parallel  sides  for 
bases,  respectively.  Fig 

Proof:  Draw  the  diag- 
onal CB.  The  triangle  ABC  has  base  AB  and  altitude 
equal  to  BE,  and  the  triangle  CBD  has  base  CD -and 
altitude  BE. 

Area  triangle  ABC  =  J    AB  X  BE. 

Area  triangle  CBD  =  j    CD  X  BE. 

Add;       trapezoid  ABDC  =  J  (AB  +  CD)  X  BE. 

164.  Cor.  The  area  of  a  trapezoid  equals  the  product  of 
the  altitude  by  a  line  joining  the  middle  points  of  the  two 
non-parallel  sides.  Prove  it. 


Proposition  V. 

165.  The  areas  of  two  triangles  having  an  equal  angle 
are  to  each  other  as  the  product  of  the  sides  including  the 
equal  angles. 

Statement:  In  the  triangles  HKL  and  MNO  let  Z  H  = 
Z  M,  then 

area  HKL  _    HK  X  HL 
area  MNO       MN  X  MO  ' 

Analysis:  Since  these  triangles  have  an  equal  angle,  it 
is  immediately  suggested  to  superpose  them  for  compari- 
son, since  we  know  that  the  two  sides,  including  the  equal 
angles  in  each,  will  take  the  same  position.  The  next 


io6  Plane  Geometry. 

suggestion  is  just  as  clear:  to  draw  a  connecting  line 
between  the  non-coincident  sides.  These  two  steps  are 
clearly  suggested  in  all  similar  cases. 

Proof:   Place  vertex  M  on  H,  making  MN  fall  on  HK, 
then  MO  will  fall  on  HL  (Why?),  and  triangle  MNO  will 

M 


Fig.  93- 

take  the  position  HNO.  Connect  N  and  L  (or  O  and  K) 
by  straight  line.  Then  HNL  is  a  triangle  intermediate 
between  HNO  and  HKL. 

Triangles  HNO  and  HNL  have  the  same  altitude   (a 
perpendicular  from  N  to  HL,  taking  HO  and  HL  as  bases). 
.-.      5NL_Hk 
HNO       HO 

Likewise  triangles  HNL  and  HKL  have  same  altitude 
(a  perpendicular  from  L  to  HK,  taking  HN  and  HK  as 

bases)  . 

HKL       HK 


HNL      HN 

Multiply  (a)  by  (b). 

T*NJ,      HKL_  HL      HK 
HNO  X-HK    ~  HO  X  HN 

(sinceHO  = 


/,  >. 


MN  by  superposition). 


Plane  Geometry.  107 

166.   Cor.  I.   Similar  triangles  are  to  each  other  as  the 
squares  of  any  two  homologous  sides. 


Fig.  94. 


Let  triangle  123  ^  triangle  456  [Z     =  similar  to.] 

T  o  •>  T  ' 

Then 


123     =     12*  =     I£ '  =  _23 

456   ~~^f  ~~~^62~~76 

12  X    13          12          13 


For     HI  = i  =  —  x  -*  (since  being  similar  their 

456       45  X  46      45      46 

angles  are  equal), 

but  -  =  -<:•  •  (why?) 

45       46 

123   _    12          £2   _      I22 


Likewise,      Hi  ==  Ii2<£  =  L3  X  f  =  etc. 

456        46  X  65        46       65        652 

167.  Cor.  77.  Similar  polygons  are  to  each  other  as 
the  squares  of  any  two  homologous  sides.  Prove  it. 

Suggestion:  Divide  the  polygon  up  into  triangles  by 
diagonals  from  any  vertex.  These  triangles  will  be  simi- 
lar. Apply  principle  of  continued  proportion. 


io8  Plane  Geometry. 

REGULAR    POLYGONS    AND    CIRCLES. 

168.  Definition:  A  regular  polygon  is  a  polygon  which 
is  equilateral  and  equiangular. 

Evidently  an  equilateral  polygon  inscribed  in  a  circle  is 
a  regular  polygon;  so  that  every  polygon,  formed  by  divid- 
ing a  circumference  into  equal  parts,  and  drawing  the  chords 
of  these  equal  arcs,  is  regular. 

Proposition    I. 

169.  Regular  polygons  of  the  same  number  of  sides  are 
similar. 

Analysis:  Since  the  sum  of  the  interior  angles  of  a  poly- 
gon depends  entirely  upon  the  number  of  sides  (What 
is  the  sum  of  the  interior  angles  of  a  polygon  of  n  sides?), 
these  polygons  will  have  the  same  total  of  interior  angles, 
and  since  they  are  regular  the  individual  angles  will  be 
equal.  Hence  they  have  equal  angles.  The  polygons 
being  regular  are  also  equilateral;  hence  all  sides  have 
the  same  ratio.  Draw  figure  and  make  proof  definite. 

170.  Cor.     Since  regular  polygons  of  the  same  number 
of  sides  are  similar,  their  perimeters  have  the  same  ratio 
as   any   two   homologous   lines;    hence   the   perimeters   of 
regular  polygons  of  the  same  number  of  sides,  are  to  each 
other  as  the  radii  of  the  inscribed  circles,  or  as  the  radii 
of  the  circumscribed  circles. 

Proposition  II. 

171.  Two  circumferences  have  the  same  ratio  as  their 
radii. 

Statement:  Let  the  circles  A  and  B  have  radii  R  and  R' 
respectively,  then  circumference  of  A  :  circumference  of 
B  :  :  R  :  R'. 


Plane  Geometry. 


109 


Analysis:  The  proportion  suggests  the  relation  between 
the  perimeters  of  regular  polygons  of  the  same  number  of 
sides,  and  regular  polygons  of  the  same  number  of  sides 
may  be  inscribed  in  both  circles.  The  number  of  sides 
may  be  infinite  without  altering  the  relation. 

Proof:  Inscribe  the  regular  polygon  of  perimeter  P  in  A 
and  the  regular  polygon  of  perimeter  P'  of  same  number 
of  sides  in  B,  then  P  :  P'  :  :  R  :  R',  however  many  sides 
P  and  P'  may  contain. 


Fig.  95- 

If  the  number  of  sides  be  infinitely  great,  the  perimeters 
of  the  polygons  coincide  with  the  circumferences  of  the 
circles,  hence  eventually  C  :  C/  :  :  R  :  R'  (where  C  and  C' 
represent  the  circumferences  of  A  and  B  respectively). 

Cor:  The  ratio  of  the  circumference  of  every  circle  to 
its  diameter  is  the  same. 

For  let  C  and  C'  be  the  circumference  of  any  two  circles 
whose  radii  are  R  and  R', 
then,  C  :  C'  :  :  R  :  R', 

C  :  C'  :  :  2  R  :  2  R'. 
C'  :  2  R', 

_____     or      C 
2R 


or 


By  alternation,    C  :  2  R 

r« 

that  is, 


2  R' 


or       -  = 


D       D' 


no  Plane  Geometry. 

It  is  customary  to  call  this  ratio  n. 
Hence,  —  =  n ;      C  =  2  x 


Proposition   III. 

172.  The  area  of  a  regular  polygon  is  equal  to  half  the 
product  of  its  perimeter  and  apothem  (apothem  meaning  the 
perpendicular  from  the  centre  upon  any  side). 

Statement:  Call  P  the  perimeter  of  a  regular  polygon  and 
h  its  apothem,  then  its  area  A  =  \  h  X  P. 

Analysis:  Any  regular  polygon  can  be  divided  into  as 
many  equal  isosceles  triangles  as  it  has  sides  by  drawing 
radii  from  the  centre  to  the  vertices.  The  apothem  is  the 
common  altitude  of  these  triangles,  and  the  sides  of  the 
polygon  are  the  bases  of  the  triangles. 

Complete  the  proof. 

Proposition  IV. 

173.  The  area  of  a  circle  is  equal  to  half  the  product  of 
its  radius  by  its  circumference. 

Statement:  Let  A  be  the  area,  R  the  radius  and  C  the 
circumference  of  the  circle,  then  A  =  J  R  X  C. 

Analysis:  Inscribe  a  regular  polygon  and  apply  Prop.  III. 

Proof:  Inscribe  a  regular  polygon  with  perimeter  P  and 
apothem  h.  Then  the  area  of  the  polygon  A'  =  i  hXP, 
no  matter  how  many  sides  it  may  have. 

If  the  number  of  sides  is  sufficiently  increased,  the  peri- 
meter of  the  polygon  approaches  the  circumference,  that 
is,  P  approaches  C  and  h  approaches  R;  hence  ultimately 
A  =  i  R  X  C  =  1 R  X  2  TrR  =  TT  R2. 


Plane  Geometry. 


in 


Proposition  V. 

174.  Given  the  side  of  a 
regular  inscribed  polygon  and 
the  radius  of  the  circle  to 
find  the  side  of  the  regular 
inscribed  polygon  of  double 
the  number  of  sides. 

Statement:  Let  AB  be  the 
given  side  of  an  inscribed 
regular  polygon  of  n  sides, 
AC  the  side  of  a  regular  in-  Fig.  96. 

scribed  polygon  of  2  n  sides,  R  the  radius  and  E  the  centre. 
Call  AB,  x.  To  find  AC  in  terms  of  R  and  x  draw  di- 
ameter CF  _L  to  AB,  intersecting  AB  at  D.  Then  D  is 
the  middle  of  AB  (Why?),  and  hence  AD  =  J  x ;  connect 
A  and  F. 

Solution: 


AE2  -  (i*)2  or  DE  =  \/R2-  -• 

*  4 

CD  =  R  -  DE  =  R  -  V/R2  -  — ; 
V  4 

also,  CF  :  AC  :  :  AC  :  CD  or  AC2  =  CF  .  CD.  (Why?) 
Observe  that  CAF  is  a  right  triangle  and  AD  a  perpen- 
dicular on  the  hypotenuse). 


=   2R 


/R-  ^/R2  -  - 


AC   == 


==  \/  2  R  /R 


-        R2  - 


=  V  R    2  R  - 


ii2  Plane  Geometry. 


If  R  =  -i  for  simplicity,  AC  =  V  2  -  \/4  -  jc2. 

176.  Cor.  To  compute  n.  Take  R  -  i  then  C  =  2  n 
or  TT  =  i  C. 

In  the  formula  let  AB  be  a  side  of  a  regular  hexagon, 
hence,  AB  =  R  =  i  Say  AB  =  Sfl,  AC  =  S12,  etc. 

Then  S12  =      v  2  -  \/4  -  i.  .51763809  +; 

hence  P12  =  6.21165708  -f 


S24  =       2  -  V  4  -  .51763809  -    .26105238  +  ; 
hence  P24  =  6.26525722  + 


S48  =  V  2  -  V  4  -  .26105238  =    .13080626  +  ; 
hence  P48  =  6.27870041  + 


Sae  =       2  -  V  4  -  .13080626  =     .06543817  +  ; 
hence  Pgg  =  6.28206396  + 

$192  =  V/2  -  ^  4  -  -06543817  =     .03272346  +  ; 
hence  P192  —  6.28290510  + 


-  -03272346  =     .01636228 
hence  P384  =  6.28311544 


^768  =  V  2  —  VX4  —  .01636228  =   .00818121  +; 
hence  P768  =  6.28316941  + 

It  is  plain  that  there  has  been  no  change  to  four  places 
of  decimals  between  the  perimeters  of  the  polygons  of 
384  and  768  sides,  hence  in  taking  these  values  for  the 
circumference  of  the  circle  there  will  occur  no  error  in  the 
first  four  places  of  decimals,  which  is  sufficiently  accurate 
for  ordinary  purposes. 

Hence,      n  =  i  C  =  J  (6.28316941  +)  =  3-14158  + 
or  71  =  3.1416  approximately. 


Plane  Geometry.  113 

EXERCISE  V. 

Areas. 

1.  Show  that  the  medians  of  a  triangle  divide  it  into  six 
equivalent  triangles. 

2.  The  area  of  a  triangle  equals  one-half  the  product  of 
its  perimeter  by  the  radius  of  the  inscribed  circle. 

3.  The   triangle   formed   by  drawing   a   line   from   any 
vertex  of  a  parallelogram  to  the  middle  of  the  opposite 
side  is  equivalent  to  J  of  the  parallelogram. 

4.  The  line  joining  the  middle  points  of  two  adjacent 
sides  of  a  parallelogram  forms  with  them  a  triangle  equal 
to  J  of  the  parallelogram. 

5.  Construct  a  square  equal  to  the  sum  of  two  given 
squares. 

6.  Equal  to  their  difference. 

7.  Change  any  given  triangle  into  an  equivalent,  right 
triangle. 

8.  Trisect   a   parallelogram   by   lines   drawn    from    one 
vertex. 

9.  The  side  of  an  equilateral  triangle  is  6.     Find  its 
area. 

10.  Show  that  the  quadrilateral  formed  by  drawing  lines 
from  the  middle  points  of  two  of  the  sides  of  a  triangle  to 
the  middle  of  the  third  side  is  a  parallelogram  whose  area 
is  half  that  of  the  triangle. 

11.  Transform  a  given  square  into  a  rectangle  having 
a  given  diagonal. 

12.  Bisect  a  trapezoid  by  a  line  parallel  to  the  base. 

13.  The  sides  of  a  triangle  are  8  in.,  12  in.,  and  6  in. 
Find  its  area. 

14.  Find  the  medians  of  the  triangle  in  Ex.  XIII. 

15.  The  sides  of  a  triangle  are      in.,  12  in.,  and  13  in. 


ii4  Plane  Geometry. 

If  a  perpendicular  is  dropped  from  the  angle  opposite  on 
the  side  13  in.,  find  the  segments  of  this  side. 

16.  The  base  of  an  isosceles  triangle  is  34  ft.,  and  one 
of  the  legs  is  15  ft.     Find  the  altitude. 

17.  The  parallel  sides  of  a  trapezoid  are  12  in.  and  10  in. 
respectively,  and  the  other  two  sides  are  9  in.  and  7  in.    Find 
the  area. 

1 8.  Construct  \/2    and   J  \/$. 

19.  The  area  of  an  equilateral  triangle  is  9  \/3-     Find 
its  sides. 

20.  The  diagonal  of  a  quadrilateral  is  24  in.  and  the 
distance  between  two  lines  parallel  to  this  diagonal  drawn 
through  the  opposite  vertices  is  10  in.     Find  the  area. 

21.  The  diagonals  of  a  parallelogram  are  8  in.  and  14  in., 
and  one  side  is  3  in.     Find  the  area. 

22.  The  sides  of  a  triangle  are  n  in.,  13  in.,  and  15  in. 
Find  the  radius  of  the  circumscribed  circle. 

23.  Two  sides  of  a  parallelogram  are  8  ft.  and  6  ft.,  and 
one  diagonal  is  14  ft.     Find  the  other  diagonal. 

24.  Two  triangles  of  equal  area  have  one  angle  in  each 
equal  and  the  sides  including  this  angle  in  one  area  are  12  in. 
and  9  in.;  one  of  the  including  sides  in  the  other  is  6  in. 
Find  the  remaining  including  side. 

25.  A  chord  10  in.  long  is  12  in.  from  the  centre  of  a 
circle.     What  is  the  area  of  the  circle? 

26.  The  radius  of  a  circle  circumscribed  about  a  regular 
polygon  is  10  in.;  the  side  of  a  polygon  is  8  in.     What  is 
the  area  of  the  polygon? 

27.  Two  tangents  are  drawn  from  a  point  to  a  circle  of 
radius  6  in.;  the  tangents  are  12  in.  long.     How  far  is  the 
point  from  the  centre  of  the  circle? 

28.  Tangents    22    in.    long    are    drawn    to   circle  hav- 


Plane  Geometry.  115 

ing  a  10  in.  radius.     Find   the   length    of    the   chord  of 
contact. 

29.  The  sides  of  a  triangle  are  21  ft.,  18  ft.,  and  14  ft. 
Find  its  area  and  from  it  the  three  altitudes. 

30.  The  area  of  a  trapezoid  is  2,150  sq.  yds.     One  of  its 
parallel  sides  is  250  yds.,  and  its  altitude  is  27  ft.     What 
is  the  other  parallel  side? 

31.  One  diagonal  of  a  rhombus  is  §  the  other,  and   their 
difference  is  6  ft.     What  is  the  area? 

32.  A  circle  whose  radius  is  6  in.  is  inscribed  in  a  quad- 
rilateral whose  perimeter  is  100  ft.     Find  the  area  of  the 
quadrilateral. 

33.  The  sides  of  a  triangle  are  in  the  ratio  2:3:4,  and 
its  area  is  108  A/I  5.     Find  the  sides. 

34.  A  circular  walk  is  242  ft.  long  and  i  yd.  wide.     How 
many  square  yards  of  cement  does  it  contain? 

35.  Find  the  area  of  a  regular  hexagon  inscribed  in  a 
circle  whose  diameter  is  10  in. 

36.  A  wheel  of  12  J  in.  radius  has  a  hole  3  in.  in  diameter 
bored  in  it  for  a  shaft.     What  is  the  area  of  the  ring  so  left  ? 

37.  Circles,  each  with  a  radius  of  7  ft.,  are  drawn  with 
the  four  vertices  of  a  square  as  centres  and  tangent  to  each 
other.     What  is  the  area  of  a  circle  tangent  to  them  all  ? 

38.  The  circumference  of  a  circle  is  33  in.     What  is  the 
side  of  an  equilateral  triangle  of  the  same  area  [n  =  3}]  ? 


SOLID    GEOMETRY 


DEFINITIONS. 

1.  A   Surface  has  extent  in  more  than  one  direction 
but  no  thickness.     As  such,  it  is  a  pure  abstraction  and 
cannot  be  concretely  represented  alone.     Give  a  concrete 
example  of  a  surface  ? 

2.  A  Plane  is  such  a  surface  that  if  any  two  of  its  points 
be  joined  by  a  straight  line  the  line  will  lie  entirely  in  the 
surface.     Give  an  example  of  a  plane  ? 

3.  A  plane  is  said  to  be  determined  by  a  stated  condition, 
when  no  other  plane  can  fulfill  that  condition. 

Cor.  I.  A  plane  cannot  be  determined  by  one  straight 
line.  (Why?) 

Cor.  II.  A  plane  can  be  determined  in  four  ways: 
By  three  points  not  in  a  straight  line;  by  a  line  and  a 
point  without  it;  by  two  intersecting  straight  lines,  and 
by  two  parallel  straight  lines.  All  four,  however,  prac- 
tically reduce  to  the  first,  for  two  of  the  points  may  be  joined 
to  furnish  the  second  condition;  or  straight  lines  may  be 
drawn  through  any  two  of  the  points,  intersecting  in  the 
third;  or  a  straight  line  may  be  drawn  through  any  one 
parallel  to  a  straight  line  joining  the  other  two. 

NOTE.  — A  plane  is  unlimited  in  extent,  but  from  Definition  No.  2 
it  is  clear  that  any  qualities  possessed  by  a  part  of  the  plane  will  be 
characteristic  of  the  whole,  hence  it  is  customary,  in  order  to  con- 
centrate and  fix  attention,  to  represent  planes  by  parallelograms  or 
rectangles. 

117 


u8 


Solid  Geometry. 


4.  A  plane  is  said  to  be  passed  through  given  points  or 
lines  when  it  is  made  to  contain  them. 

5.  A  line  is  said  to  be  perpendicular  to  a  plane  when  it 
is  perpendicular  to  every  (or  any)  line  of  the  plane  through 
its  foot. 

How  would  you  denote  a  line  neither  perpendicular  nor 
parallel  to  a  plane  ? 

NOTE.  —  The  point  where  a  perpendicular  pierces  the 
plane  is  called  its  foot. 

6.  A  line  is  said  to  be  parallel  to  a  plane  if  the  line  and 
plane  cannot  meet,  however  far  produced.     Give  a  defini- 
tion for  parallel  planes. 

7.  The  intersection  of  two  or  more  planes  is  the  locus 
of  all  points  common  to  the  intersecting  planes. 

8.  Points  or  lines  lying  in  the  same  plane  are  said  to  be 
coplanar.     Points    lying    on    the    same    straight    line    are 
called  collinear. 


Proposition  I. 

The  intersection  of 
two  planes  is  always 
a  straight  line. 

Analysis:  The  def- 
inition of  a  plane 
(Art.  2)  suggests  a 
method  of  proof, 
since  if  we  take  any 
two  points  in  the 
intersection  and  join 
them  by  a  straight  line,  the  definition  (Art.  7)  of  intersec- 
tion applies. 


Solid  Geometry.  up 

Concrete  statement  from  figure: 

Let  the  plane  AC  intersect  the  plane  EF.  Then  their 
intersection  is  a  straight  line. 

NOTE.  —  Observe  that  AC  and  EF  are  very  small  sec- 
tions of  the  whole  plane. 

Proof:  Take  any  two  points  in  the  intersection,  say  M 
and  N.  Join  these  points  by  a  straight  line.  Since  these 
are  points  of  the  intersection,  they  lie  in  both  planes  at 
the  same  time. 

.'.  the  line  joining  them  lies  wholly  in  both  planes.   (Why  ?) 

.'.  it  is  their  intersection.  (Why?) 

.".  their  intersection  is  a  straight  line. 

Comment:  There  are  three  general  methods  of  demon- 
stration: the  direct  method,  the  indirect  (like  above  proof), 
and  the  reduction  to  an  absurdity  (reductio  ad  absurdum). 
The  first  method  is  a  straight  forward  progression  from 
hypothesis  to  conclusion.  The  second  is  the  assumption 
that  certain  known  lines  are  drawn  according  to  the  pre- 
scribed conditions,  and  these  lines  proved  identical  with 
the  given  lines.  The  third  is  the  assumption  of  the  con- 
trary to  the  proposition,  and  the  proof  that  this  assump- 
tion leads  to  a  geometric  absurdity.  The  next  proposition 
will  illustrate  the  first  method. 

Proposition  II. 

//  a  straight  line  is  perpendicular  to  two  straight  lines  at 
their  intersection  point  it  is  perpendicular  to  their  plane. 

Analysis:  The  definition  says  that  a  perpendicular  to 
a  plane  must  be  perpendicular  to  every  line  in  the  plane 
drawn  through  its  foot.  It  is  necessary  then  to  prove  that 
the  line  that  is  perpendicular  to  two  intersecting  lines  at 
their  intersection,  is  also  perpendicular  to  any  other  line 
drawn  through  that  point. 


I2O  Solid  Geometry. 

Any  line  means,  essentially,  every  line.  If  any  other 
line  is  drawn  through  the  foot  of  this  perpendicular,  to 
prove  it  is  also  perpendicular,  probably  the  simplest  process 
is  to  show  that  it  makes  a  right  angle  with  it.  To  show 
that  it  makes  a  right  angle,  since  a  triangle  is  easily  con- 
structed, of  which  the  perpendicular  and  this  random  line 
are  sides,  it  is  suggested  to  prove  this  triangle  a  right  tri- 
angle. What  is  the  essential  relation  between  the  sides 
of  a  right  triangle  ? 

Statement:  Let  CD  and  EF  be  the  intersecting  lines  in 
the  plane  ccy,  and  AB  the  perpendicular  to  them  at  B, 
their  intersection,  then  is  AB  perpendicular  to  xy. 

Prooj:  Draw  any  line  BL 
through  B.  We  are  to  show 
that  AB  is  J_  to  BL. 

To  include  these  lines  in 
triangles  and  to  make  these 
triangles  as  useful  as  possi- 
ble, draw  GK,  cutting  CD  at 
G  and  EF  at  K,  and  inter- 
secting BL  in  such  a  way 
that  the  intersection  point  H  bisects  GK. 

(Show  how  to  draw  GK  to  fulfill  this  condition.)  Join 
A  with  G.  H  and  K.  Then  AH  is  the  median  of  the  triangle 
KAG,  and  BH  is  the  median  of  KBG. 

Hence,    AK2  +  ACP  =  2  A~H2  +  2  HK2  (i) 

[by  Plane  Geometry,  152.] 
and  BK2  +  BG2  =  2  BH2  +  2  HK2  (2) 

It  will  be  observed  that  in  these  two  equations  are  in- 
volved two  of  the  sides  (hypotenuse  and  a  leg)  of  the  two 
right  triangles  ABK  and  ABG,  so  that  if  (2)  be  subtracted 


Solid  Geometry. 


121 


from  (i)  we  will  get  in  each  case  the  difference  of  the 
squares  of  these  hypotenuses  and  sides,  and  since  AB  is 
the  missing  leg  in  each  of  these  triangles,  the  above  dif- 
ference will  get,  in  both  cases,  AB2.  Since  AB  is  a  side  of 
the  triangle  ABH,  which  we  want  to  prove  right,  the  sub- 
traction is  suggested. 


Subtract  (2)  from  (i); 
~(AK^-"BK2)  +(AG2- 


-  2  (AIT2  -  ~BlP) 


AB2  +  AB2 

2 


2    (AH2  -  BIT) 

-2 


AB'=   AH'-   BH 
.'.  ABH  is  a  right  triangle,  and  AB  is  perpendicular  to  BH. 


Proposition  III. 

Only  one  perpendicular  can  be   drawn  to  a  given  plane 
through  a  given  point. 

Analysis:  There  are  plainly  two  cases:  first,  when  the 
point  is  within  the  plane;  second,  when  it  is  without. 

First,  the  JL  must  be  drawn, 
and  by  definition  it  must 
be  _J_  to  every  line  in  that 
plane  through  the  given  point. 
But  by  Prop.  II,  it  will  now 
be  sufficient  to  make  it  _J_ 
to  two  lines  through  the 
point. 

Having  drawn  the  J_  we 
can  prove  our  proposition  by 
showing  that  any  other  line  Fig.  3. 

through  the  given  point  is  not  _L  to  the  plane. 

Proof :  Let  B  be  the  given  point  in  the  given  plane  MN. 
Prove  that  One  perpendicular  alone  can  be  drawn  at  this 


122 


Solid  Geometry. 


point  to  plane  MN.  Through  B  draw  any  line  CD  in  the 
plane,  and  through  B  pass  a  plane  RS  J_  to  CD. 

In  the  plane  RS  draw  AB  J_  to  FK,  the  intersection  of 
RS  and  MN;  AB  is  thej_  required.  For  CD  isj_  to  plane 
RS  by  construction,  hence  CD  is  J_  to  AB  a  line  in  RS 
through  foot  of  CD;  that  is,  reciprocally,  AB  is_L  to  CD. 
But  AB  was  drawn  J_  to  FK  which  intersects  CD  at  B  and 
lies  in  the  same  plane.  .'.  by  Prop.  II,  AB  is  J_  to  plane 
MN  of  FK  and  CD.  Draw  any  other  line,  say  EB,  through 
B  to  show  that  it  is  notj_  to  MN. 

Since  EB  and  AB  intersect  in  B,  they  determine  a  plane; 
that  is,  they  lie  in  the  same  plane  (this  plane  may  be  RS  or 
some  other,  depending  upon  the  way  EB  is  drawn) .  Now 
if  EB  is_L  to  MN  through  B,  it  isj_  to  CD.  But  AB  is 
already  proved  J_  to  CD,  which  would  give  us  two  J_'s  to 
CD  at  the  same  point  lying  in  the  same  plane,  but  by 
plane  geometry  that  is  impossible :  therefore  EB  is  not  J_  to 
MN,  and  since  it,  being  any  line  but  AB,  represents  every 
line  through  B,  except  AB,  there  is  no  other  J_  to  MN 
through  B. 

If  point  is  without  the  plane, 
the  proof  must  be  somewhat 
altered,  as  there  is  not  now  a 
definite  point  in  the  plane 
through  which  to  draw  the 
random  line. 

Statement :    From  A   without 
the  plane  RS  to   draw  a  _L  and 
to  show  that  it  is   the  only  one 
from  A  to   RS.      In    RS    draw 
Fie-4.  any   line    FG,    and    through    A 

draw  the  plane  MN  _L  to  FG;  it  will  intersect  RS  in 
xy.  From  A  drop  a  _L  AD  in  MN  to  xy.  It  is  the  JL 
required. 


Solid  Geometry. 


123 


Proof:  Through  D  draw  any  line  DC  intersecting  FG 
at  C.  Prolong  AD  to  E,  making  DE  equal  to  AD.  Con- 
nect A  and  E  with  C  and  B,  the  points  in  which  FG  inter- 
sects DC  and  xy  respectively. 

In  the  triangles  ABC  and  EBC,  Z  ABC  and  Z  EEC 
are  right  ^,  for  AB  and  BE  are  in  the  plane  RS  to  which 
FG  is_J_  by  construction,  hence,  FG  isj_  to  AB  and  BE 
(definition).  Also  BC  is  common  to  both  triangles  and 
AB  =  BE,  because  xy  isj_  to  AE  at  its  middle  point  D. 
/.  the  two  right  triangles  are  equal  and  AC  =  EC.  .'.  DC 
is  J_  to  AE  at  its  middle  point,  that  is,  AE  or  a  part  of 
AE,  AD  isj_  DC,  and  it  was  drawn  J_  to  xy  or  DB. 
.'.  AD  isj_  to  the  plane  of  xy  and  DC,  that  is,  to  RS. 

Now  to  prove  that  it  is  the  onlyj_,  the  analysis  is  the 
same  as  before,  and  the  proof  practically  the  same.  Give  it. 


Proposition  IV. 

Oblique  lines  drawn  from  any  point  of  a  perpendicular 
to  a  plane,  meeting  it  at  equal 
distances  from  the  foot  of  the 
perpendicular,  are  equal;  of  two 
oblique  lines  meeting  the  plane 
at  unequal  distances  from  the  foot 
of  the  perpendicular  the  more  dis- 
tant is  the  greater. 

Statement:  Let  P  be  a  point  on 
the   _L    OX,    SP,    NP    and    MP 
oblique  lines  meeting    the    plane 
AC  so  that  ON  =   OM,  but  OS   >    ON  or  OM;  then  is 
NP  ==  MP  and  SP  >  NP  or  MP. 

Analysis:  It  is  evident  that  the  three  lines,  the  J_,  the 


124  Solid  Geometry. 

oblique,  and  the  distance-lines,  as  ON,  OM  and  OS,  form 
right  triangles.  Why  right  triangles  ? 

It  is  then  only  necessary  to  prove  equality  or  inequality 
of  the  triangles  to  establish  our' proposition. 

Proof  (ist  part):  In  the  right  triangles  PON  and  POM, 
PO  is  common  and  ON  =  OM  (by  hypothesis).  .'.  triangle 
PON  =  triangle  POM.  .'.  PN  =  PM. 

Proof  (2d  part):  On  OS  lay  off  OR  =  ON  =  OM; 
draw  RP.  Then  by  ist  part  RP  =  NP  =  MP.  But  ORP 
and  OSP  being  now  in  the  same  plane,  our  demonstration 
reduces  to  the  Plane  Geometry  Proposition:  If  from  any 
point  of  a  line  oblique  lines  are  drawn,  etc.  (complete  the 
statement  of  the  proposition).  .'.  SP  >  RP.  But  RP  - 
NP  =  MP.  .'.  SP  >  NP  or  MP. 

Cor.  I.  If  two  equal  oblique  lines  are  drawn  from  any 
point  in  a  perpendicular  to  a  plane,  they  meet  the  plane  at 
equal  distances  from  the  foot  of  the  perpendicular. 

Cor.  II.  Since  triangles  ORP,  OMP,  and  ONP  are  equal, 
the  A  ORP,  ONP,  and  OMP  are  equal.  But  Z  ORP 
>  Z  OSP  (Why?),  and  since  SP  >  RP,  the  oblique  line 
having  the  less  inclination  to  a  plane  is  the  greater. 

Definition:  The  projection  of  a 
line  on  a  plane  is  the  line  of  the 
plane  included  between  the  feet  of 
the  perpendiculars  dropped  from 
each  extremity  of  the  line.  If  the 
line  touches  the  plane,  its  projec- 
tion is  the  distance  in  the  plane, 
V  from  the  point  of  intersection  to 
the  foot  of  the  perpendicular 

dropped  from  the  other  end  of  the  line,  thus,  ED  is  the  pro- 
jection of  BC  on  plane  xy,  and  AD  is  the  projection  of  AC 
on  same  plane,  A  being  intersection  point  of  AC  and  xy. 


Solid  Geometry.  125 

Proposition  V. 

The  locus  of  points  in  space  equidistant  from  the  ex- 
tremities of  a  straight  line  is  the  plane  perpendicular  to 
this  line  at  its  middle  point. 

Statement:  The  locus  of  points  equally  distant  from  the 
extremities  of  AB  is  the  plane  MN  J_  to  AB  at  its  middle 
point  O.  A 

Analysis:  It  is  necessary,  as 
in  the  case  of  all  loci,  to  prove 
that  all  points  in  MN  are 
equally  distant  from  A  and 
B,  and  also  that  any  point 
outside  of  MN  is  unequally 
distant. 

Proof:  Let  C  be  any  point 
in  MN.  Join  O  and  C,  and  Fig>  7' 

C  with  A  and  B.  Then  AOC  and  OCB  are  right  triangles 
(Why?),  having  OC  common  and  AO  =  OB  by  construc- 
tion, hence  triangle  AOC  =  triangle  OCB,  and  .'.  CB  = 
AC,  that  is,  C  is  equally  distant  from  A  and  B. 

2d:  Let  C'  be  any  point  not  in  MN.  Join  C'  with  A  and 
B,  also  D  (where  BC'  cuts  the  plane)  with  A. 

AD  =  DB  by  first  part. 

C'D  +  AD  >  AC',  or,  since  AD  -  DB,  C'D  +  DB  >  AC', 
that  is,  BC'  >  AC'. 

C'  being  any  point  outside  of  the  plane  MN,  every  point 
outside  is  unequally  distant. 

Proposition  VI. 

All  straight  lines  perpendicular  to  a  given  straight  line  at 
a  given  point  lie  in  a  plane  perpendicular  to  the  given  line 
at  the  given  point.  How  else  may  this  proposition  be  stated  ? 


126 


Solid  Geometry. 


Fig.  8. 


Statement:  Let  BC  be  any  perpendicular  to  EB  at  B. 
To  show  that  BC  lies  in  the  plane  MN  _[_  to  EB  at  B. 

Analysis:  If  it  can  be  shown 
that  such  a  plane  is  determined 
by  these  perpendiculars  the 
proposition  is  established. 

Proof:  Draw  any  other  per- 
pendicular to  EB  at  B,  as  DB. 
Then  BC  and  DB,  being  inter- 
secting lines,  determine  a  plane, 
say  MN.  Since  EB  is  J_  to  both 
BC  and  BD  at  their  intersection, 
it  is  J_  to  their  plane  MN.  Hence,  inversely,  MN  is  J_  to 
EB  at  B. 

BC  and  DB  being  any  lines  represent  every  line,  hence 
MN  is  the  locus  of  lines  J_  to  EB  at  B.     Is  this  complete? 

Proposition  VII. 

Two  straight  lines  perpendicular  to  the  same  plane  are 
parallel. 

Statement :  Let  AB  and  CD  be 

Analysis:  To  be  parallel  they 
must  clearly  lie  in  one  plane, 
and  if  in  that  plane  they  are 
both  J_  to  the  same  straight 
line,  they  are  ||. 

Proof:  Join  B  and  D,  the 
feet  of  the  two  perpendiculars. 
Through  D  draw  xy  J_  to  BD; 
also  lay  off  ED  =  =  ED,  and 
connect  A  with  F,  D  and  E,  and  B  with  F  and  E. 

BF  =  BE  (Why?),  hence  AE  =  AF  (Why?).  /.  AD  is 
J_  to  xy  at  D  (Why?).     Hence  EF  is  J_  to  AD  and  BD  at 


to  MN,  then  they  are 
A  C 


M 

\V 

V: 

IS 

f 

r 

D 

"^1      \    7 

sA 


Fig.  9- 


Solid  Geometry.  127 

D;  it  is  also  J_  to  CD,  for  CD  was  drawn  J_  to  plane  which 
contains  xy.  But  AB  lies  in  the  same  plane  with  AD 
and  BD  (Why?),  and  since  AD  and  BD  determine  a 
plane  with  CD,  .'.  AB  and  CD  AC* 

are  in  the  same  plane,  and  are 
both  J_  to  the  line  BD  (Why?),    x- 
hence  by  Plane  Geometry,  they 
are  ||. 

Cor.  I.  If  one  of  two  parallel 
lines  is  perpendicular  to  a  plane, 
the  other  is  also  perpendicular  to  Fi£' I0* 

the  plane.  That  is,  if  AB  and  CD  are  ||,  and  AB  is  per- 
pendicular to  MN,  CD  is  also  J_  to  MN.  Prove  it. 

Cor.  II.  If  two  straight  lines  are  parallel  to  a  third 
straight  line  in  space,  they  are  parallel  to  each  other. 

Let  AB  and  CD  be  ||  to  EF,  then  AB  and  CD  are  ||  to 
each  other.  For,  pass  a  plane  xy  JL  to  EF,  then  by  Cor.  I, 
AB  and  CD  are  both  J_  to  xy,  hence  by  Prop.  VII  they 
are  ||  to  each  other. 

Proposition  VIII. 

//  two  straight  lines  are  parallel,  every  plane  passed 
through  one  of  them  and  only  one  is  parallel  to  the  other. 

A Statement :  AD  is  ||  to 

EB.      The    plane    CG 
through  EB  is  ||  to  AD. 
Analysis:  The  proof 
plainly    depends    upon 
the  definition  of  paral- 
~kG-  lelism.     It   remains   to 
Fig-  "'  determine  whether    the 

line  AD  and  plane  CD  can  meet,  and  if  so,  how? 

Proof:  AD  and  EB  are  in  the  same  plane  (Why?),  hence 


128 


Solid  Geometry. 


if  AD  meets  the  plane  CD  it  must  meet  it  somewhere  in 
EB  ;  but  AD  cannot  meet  EB  at  all,  for  they  are  ||,  .'.  AD 
cannot  meet  CD. 

Cor.  I.  To  pass  a  plane  through  either  one  of  two  lines 
in  space,  parallel  to  the  other.  For  instance,  to  pass  a 
plane  through  CD  parallel  to  AB.  Through  any  point  of 
CD  as  E  draw  a  line  EF  ||  to  AB,  then  the  plane  deter- 
mined by  CD  and  EF  will  be  the  required  plane  (Why?). 


Fig.  12. 


Fig.  13. 


Cor.  II.  To  pass  through  a  given  point  a  plane  par- 
allel to  two  given  lines  in  space.  To  pass  a  plane  through 
the  point  R  parallel  to  AB  and  CD. 

Through  R  draw  Oz  ||  to  AB,  and  xy  \\  to  CD.  The 
plane  determined  by  xy  and  Oz  is  the  required  plane 
(Why?). 

Proposition  IX. 

//  a  straight  line  is  parallel  to  any  line  in  a  plane,  it  is 
parallel  to  the  plane. 

Statement :  Let  B  be  a  line  in  the  plane  of  which  MN  is  a 


Solid  Geometry.  129 

portion,  and  let  the  line  A  be 
parallel  to  B,  then  is  A  par- 
allel to  the  plane  of  MN.  M_ 

Analysis:  To  prove  paral- 
lelism we  must  show  an  im- 
possibility of  meeting.  We 
know  the  lines  A  and  B  to 
be  parallel,  and  if  we  can 
make  the  meeting  of  A  and  Fig.  14. 

the  plane  depend  upon  the 
meeting  of  A  and  B  we  can  establish  our  theorem. 

Proof:  A  and  B  are  in  the  same  plane  (Why?),  and  hence 
if  A  meets  the  plane  of  MN,  it  must  meet  the  line  B  (the 
lines  being  sufficiently  produced).  But  A  and  B  are  par- 
allel by  hypothesis,  and  hence  cannot  meet,  .'.  A  cannot 
meet  the  plane  of  MN. 

Cor.  I.  If  a  line  is  parallel  to  a  plane,  the  intersection 
of  that  plane  with  a  plane  through  the  line  is  parallel  to 
the  line. 


130 


Solid  Geometry. 


Proposition  X. 


The  intersection  made  by  two  parallel  planes  with  a  third 
plane  are  parallel  straight  lines. 

Statement:  If  the  two  parallel  planes  AB  and  CD  are 
cut  by  the  plane  RS  intersecting  AB  in  ocy  and  CD  in  vz, 
then  xy  is  ||  to  vz. 

Analysis :  Parallelism  is  determined  by  the  non-inter- 
section of  the  lines  which  must  be  proved  to  lie  in  the 
same  plane. 

Proof  :  XY  and  VZ  lie  in  the  same  plane  RS,  byhypoth- 


Fig.  15. 


esis,  and    also    in    the    planes  AB  and  CD  because  they 
are  the  intersection  of  these  planes  with  RS. 

Since  AB  and  CD  are  ||  by  hypothesis,  they  cannot  meet, 


Solid  Geometry. 


,'XH 


and  hence  no  lines  in  one  can  meet  any  lines  in  the  other. 
Hence,  being  in  the  same  plane  RS  and  unable  to  meet, 
XY  is  ||  to  VZ. 

Cor.  Parallel  lines  included  between  parallel  planes  are 
equal. 

Proposition  XI. 

//  two  intersecting  straight  lines  are  both  parallel  to  a 
plane,  the  plane  determined  by  these  R 
lines  is  parallel  to  that  plane. 

Statement :  If  NP  and  LM  are 
two  intersecting  lines,  both  ||  to 
the  plane  xy,  then  RS  determined 
by  NP  and  LM  is  ||  to  xy. 

Analysis:  Since  there  is  no  op- 
portunity  of  testing  directly  the 
possibility  of  the  meeting  of  xy  and 
RS,  we  must  establish  their  paral- 
lelism by  their  relation  to  some  line 
or  plane,  preferably  a  perpendicular;  hence, 

Proof:  Draw  the  line  OB  _[_  to  plane  xy  from  the  point  O 
(Why  from  O?).  Through  OB  and  NP  and  OB  and  LM 
pass  planes  PK  and  LH,  intersecting  xy  in  KF  and  GH. 
Then  GH  is  ||  to  LM  and  KF  is  ||  to  NP  (Prop.  9,  Cor.  I). 
OB  is  J_  to  KF  and  GH  at  B  (Why?),  hence  OB  is  J_  to 
NP  and  LM  at  O  (Why?);  .'.  OB  is_[_  to  plane  RS  (Why?). 
If,  then,  both  planes,  RS  and  xy,  are  J_  to  OB  and  could 
intersect,  lines  drawn  in  them  from  O  and  B  to  any  point 
of  their  intersection  would  both  be  J_  to  OB.  This  is  im- 
possible, hence  RS  and  xy  are  parallel. 

Cor.  It  was  proved  in  Prop.  XI  that  OB,  which  was 
drawn  _[_  to  plane  xy,  was  also  _L  to  RS,  which  was  finally 
proved  ||  to  xy,  hence,  the  proof  being  entirely  general,  it 


Fig.  16. 


132  Solid  Geometry. 

follows,  that  if  a  line  be  J_  to  one  of  two  ||  planes  it  is  J_  to 

the  other  also. 

Proposition  XII. 

//  two  angles,  although  in  different  planes,  have  their 
sides  parallel,  each  to  each,  they  are  equal,  and  their  planes 
are  parallel. 

Statement:  If  the  sides  AR  and  AS  of  angle  A  are  ||  to 
sides  DH  and  DK  of  angle  D  respectively,  then  Z  A  = 
Z  D,  and  plane  xy  is  ||  to  plane  MN. 
Analysis:  Where  angles   are  not 
adjacent  or  immediately  associated 
with  each  other,  it  is  necessary,  in 
order  to  compare  them,    to  make 
them  parts  of  closed  figures  (pref- 
erably   triangles)     which    can    be 
compared. 

Fig.  X7.  As   there  are   no   limits   set   for 

the  sides  of  these  angles,  we  are  free  to  lay  off  any  dis- 
tance on  them  we  choose.  Clearly,  it  is  advisable  to  make 
these  distances  equal. 

Proof:  Lay  off  AB  (on  AR)  and  DE  (on  DH)  so  that 
AB  =  DE  ;  also  AC  (on  AS)  =  DF  (on  DK),  join  B 
and  C,  and  E  and  F.  Draw  AD,  BE  and  CF,  joining 
A  and  D,  B  and  E,  and  C  and  F  (Why?).  Since  AC  is 
equal  and  parallel  to  DF,  by  hypothesis,  ACFD  is  a 
parallelogram;  for  same  reason  ADEB  is  a  O,  hence 
AD  is  equal  and  parallel  to  CF  and  BE,  hence  they  are 
equal  and  parallel  to  each  other  (Why?).  Therefore, 
BCFE  is  a  O  (Why?),  and  .'.  BC  =  EF.  Hence  the 
triangles  ABC  and  DEF  have  three  sides  equal,  each  to 
each,  .'.  the  triangles  are  equal  and  Z  A  =  Z  D. 

Since  two  intersecting  lines  determine  a  plane,  and  AR 
is  II  to  DH,  and  AS  is  ||  to  DK,  plane  xy  is  ||  to  plane  MN. 


Solid  Geometry.  i^o 

EXERCISE    I. 

1.  Show  how  to  erect  a  rod  perpendicular  to  a  plane 
surface. 

2.  Prove  that  only  one  plane  can  be  drawn  through  a 
given  straight  line  parallel  to  another  given  straight  line. 

3.  Draw  a  straight  line  parallel  to  a  given  straight  line, 
and  meeting  each  of  *two  other  straight  lines  not  situated 
in  the  same  plane. 

4.  Draw   a  straight  line   of  given   length  parallel  to  a 
given  plane,  and  having  its  extremities  on  two  given  straight 
lines  not  in  the  same  plane. 

5.  Find  the  locus  of  lines  through  a  given  point  parallel 
to  a  given  plane. 

6.  Find  the  locus  of  points  in  a  plane  equidistant  from 
two  points  not  in  the  plane  (two  cases). 

7.  Find  the  locus  of  points    in   space  equidistant  from 
'three  points  not  in  a  straight  line. 

8.  Find  a  point  in  a  given  plane  such  that  the  sum  of 
its  distances  from  two  points  on  the  same  side  of  the  plane 
shall  be  a  minimum. 

9.  What  is  the  locus  of  points  in  space  equally  distant 
from  the  vertices  of  a  triangle  ? 

10.  If  a  plane  is  passed  through  a  diagonal  of  a  par- 
allelogram,  the  perpendiculars  upon  this  plane  from  the 
ends  of  the  other  diagonal  are  equal. 

11.  Through  a  point  in  a  plane  draw  a  line  in  the  plane 
parallel  to  a  given  line  outside  the  plane. 

Proposition  XIII. 

//  two  straight  lines  are  cut  by  three  parallel  planes,  their 
corresponding  segments  are  proportional. 

Statement:  If  the  two  straight  lines  xy  and  vz  are  cut  by 


134  Solid  Geometry. 

the    three  planes  NM,   PQ  and 
M      /          \  RS    at    the  points   A,  B    and    C 

(on  xy)  and  F,  E  and  D  (on  vz), 
thenAB  :  BC  ::  FE  :  ED. 

Analysis:  As,  in  general,  it  is 
necessary  to  join  the  two  lines  in 
some  way,  say  their  extreme 
points  of  intersection  with  the 
planes  MN  and  RS  by  a  straight 
line,  which  will  plainly  give  us 
two  pairs  of  intersecting  lines 
(AD  and  AC,  and  AD  and  FD), 
to  determine  two  planes,  which 
JS  will  intersect  the  three  parallel 

planes  in  parallel  lines. 
y  Proof:    Join    A   and    D   by  a 

straight    line,    then   through   AC 

and  AD  pass  a  plane  cutting  PQ  in  BG  and  RS  in  CD? 
also  through  AD  and  FD  a  plane,  cutting  PQ  in  GE  and 
MN  in  AF.     In  the  triangle  ACD,  BG  is  ||  to  CD  (Why?) 
/.  AB  :BC  :  :  AG  :  GD.     (i) 

In  the  triangle  DAF,  GE  is  ||  to  AF.  (Why?) 

.'.  DG  :  GA  :  :  DE  :  EF,   or    (by  inversion)  AG  :  GD 
::FE:ED.     (2) 
Combining  (i)  and  (2);  AB  :  BC  :  :  FE  :  ED. 

Definitions. 

The  space  between  two  intersecting  planes  is  called  a 
dihedral  angle.  The  amount  of  divergence  of  the  two  planes 
determines  the  size  of  the  angle. 

The  two  planes  are  called  the  faces  of  the  dihedral  angle, 
and  their  intersection  its  edge. 


Solid  Geometry. 


135 


Dihedral  angles  are  called  adjacent,  right,  oblique,  etc., 
under  the  same  circumstances  as  plane  angles,  planes 
replacing  lines  in  the  definitions. 

Dihedral  angles  are  estimated  and  compared  by  their 
plane  angles,  which  are  formed  by  drawing,  in  each  face, 
straight  lines  perpendicular  at  the  same  point  to  the  edge. 

Many  of  the  propositions  referring  to  plane  angles  apply 
directly  to  dihedral  angles;  as, 

(a)  If   two  planes   intersect,  either  one   forms  with  the 
other  two  adjacent  angles,  whose  sum  is  two  right  dihedral 
angles. 

(b)  If  two  planes  intersect,  the  vertical  dihedral  angles 
are  equal. 

(c)  If  two  parallel  planes  are  cut  by  a  transversal  plane, 
the  alternate  angles  and  the  interior-exterior  angles  on  the 
same  side  of  the  transversal  are  equal;  also  the  two  interior 
dihedral  angles  on  the  same  side  of  the  transversal  are 
supplementary,  etc. 

Proposition  XIV. 

Two  dihedral  angles  are  equal  if  their  plane  angles  are 
equal. 

Statement:  The  two  dihedral  angles  BC  and  FH  (dihe- 
dral angles  are  designated  by  their  edges  when  standing 
alone;  thus,  A-BC-D,  p 
E-FH-G,  if  not)  are 
equal  if  the  plane  angles 
xyz  and  123  are  equal. 

Analysis:  Here,  as  in 
Plane  Geometry,  the  final 
test  of  equality  is  coinci- 
dence, when  one  figure  is 
applied  to  another. 

Proof:  Apply  the  edge 


136 


Solid  Geometry. 


of  FH  to  the  edge  of  BC,  so  that  the  point  2  falls  on 
point  y,  and  the  plane  EH  falls  on  plane  AC  ;  then  12 
will  fall  on  yz,  and  since  Z_  ocyz  =  Z  123  (by  hypothesis) 
23  will  fall  on  yz;  also  FH  coinciding  with  BC  (two  inter- 
secting lines  determine  a  plane),  plane  FG  coincides  with 
plane  BD,  hence  angle  BC  coincides  throughout  with  FH. 


Proposition  XV. 

Two  dihedral  angles  have  the  same  ratio  as  their  plane 
angle. 

Statement:  The  two  dihedral  angles  AB  and  MN  have 
the  plane  angles  CAG  and  OMX. 

Analysis:  There  are  clearly  two  cases:  one,  when  the 
two  angles  have  a  common  measure,  which  they  contain, 
each,  an  integral  number  of  times;  another,  when  they 
have  no  common  measure,  that  is,  are  what  is  called  in- 
commensurable. 

In  the  first  case,  we  can  compare  them  by  comparing 
the  number  of  times  each  takes  the  common  measure. 
In  the  second  case,  we  can  use  the  same  method,  if  we 
G  take    a    unit    of   measure    so 

small  that  it  leaves  an  insig- 
nificant remainder. 

Proof:  Case  I.  Divide  the 
angles  CAG  and  OMX  into 
parts  each  equal  to  the  unit 
of  measure  that  they  will  con- 
tain; say  it  is  contained  in 
CAG,  m  times,  and  in  OMX, 
:  m  :  n.  (i) 


B 

Fig.   20. 

n  times;  then  CAG  :  OMX 


Through   these  points   of  division   and   the   edges   AB 
and  MN  pass  planes;  these  planes  will  divide  the  dihedral 


Solid  Geometry. 


Z  ABD  :  Z  OMP    when 


angles  AB  and  MN  respectively  into  m  and  n  equal  dihe- 
dral angles;  hence,  Z  AB  :  Z  MN  :  :  m  :  n.  (2) 

Combining  proportions  (i)  and  (2)  having  equal  couplets, 
Z  AB  :  Z  MN  :  :  Z  CAG  :  Z  OMX. 

Proof:  Case  II. 

Statement:    Z  BK  :  Z  MT  : 
the    angles   are   incommen- 
surable. 

Proof:  If  we  take  an 
angle  unit  exactly  con- 
tained in  Z  OMP,  say,  and 
divide  ABD  by  that  unit, 
it  will  be  contained  in  Tl  u 

Z  ABD  a  certain  number  of 
times  with  a  remainder  FBD.  -  Fig.  21. 

Planes  passed  through  these  points  of  division  will  divide 
the  dihedral  angles  into  corresponding  parts,  Z  MT  into 
an  integral  number  of  parts,  and  BK  into  a  number  of 
parts  with  a  remainder  F-BK-D,  which  of  course  will  be 
less  than  the  unit  used. 

Then  the  part  A-BK-F  of  BK  is  commensurable  with  MT, 
and  hence, bypart  i,Z  A-BK-F  :  Z  MT  :  :Z  ABF  :  Z  OMP. 

Now  by  taking  a  smaller  unit,  say  an  aliquot  part  of 
the  original  unit,  the  Z  MT  will  still  contain  an  integral 
number  of  parts;  and  the  remainder  in  BK,  since  it  must 
always  be  less  than  the  dividing  unit,  can  be  made  smaller 
and  smaller,  as  we  use  a  smaller  and  smaller  unit.  Clearly, 
there  is  no  limit  to  the  decrease  in  the  remainder,  hence 
we  can  make  the  remainder  as  nearly  zero  as  we  please;  but 
wherever  the  division  point  F  is  situated,  the  proportion 
Z  A-BK-F  :  Z  MT  :  :  Z  ABF  :  Z  OMP  is  always  true, 
even  when  the  point  F  is  practically  coincident  with  the  point 
D;  hence,  in  general,  Z  BK  :  Z  MT  :  :  Z  ABD  :  Z  OMP. 


138  Solid  Geometry. 

Proposition  XVI. 

//  two  planes  are  perpendicular  to  each  other,  any  line 
drawn  in  one  of  them,  perpendicular  to  their  intersection,  is 
perpendicular  to  the  other  plane. 


Fig.  22. 

Statement :  If  the  planes  MN  and  xT  are  J_  to  each 
other,  the  line  PQ  drawn  in  xT,  J_  to  zy,  the  intersection 
of  the  planes  is  _[_  to  MN. 

Analysis:  Since  the  plane  angle  of  a  dihedral  measures 
it  and  in  this  case  PQ  is  one  side  of  such  angle,  it  is 
plainly  indicated  to  complete  this  angle.  Hence, 

Proof:  In  MN  draw  QR  _L  to  zy  at  Q.  Then  since 
PQ  and  QR  are  bothj_  to  zy  by  construction,  Z  PQR 
is  the  plane  angle  of  the  dihedral  having  zy  for  its  edge, 
but  this  dihedral  angle  is  a  right  angle  by  hypothesis. 
.'.  Z  PQR  is  a  right  angle;  that  is,  PQ  is_[_  to  QR,  but 
it  is  also  _L  to  zy;  .'.  it  is  J_  to  MN.  (Why?) 


Solid  Geometry.  139 

EXERCISE. 

Prove  that  if  two  planes  are  perpendicular  to  each  other, 
a  perpendicular  to  one  of  them  at  a  point  in  their  inter- 
section will  lie  in  the  other. 

Suggestion:  Draw  a  J_  in  the  other  plane  and  prove  its 
identity  with  the  given  J_. 

Cor.  If  two  planes  are  perpendicular  to  each  other,  a 
perpendicular  to  one  of  them  from  any  point  in  the  other 
will  lie  in  that  other  plane. 

Proposition  XVII. 

//  a  straight  line  is  perpendicular  to  a  plane,  every  plane 
passed  through  that  line  will  be  perpendicular  to  the  plane. 


Fig.  23. 

Statement:  If  the  line  MO  is_j_  to  plane  LK,  intersecting 
it  at  O,  and  the  plane  AB  be  any  plane  passed  through  MO 
AB  is  JL  to  LK. 

Analysis:  Construction  of  plane  angle,  evidently. 

Proof:  In  LK  draw  to  the  point  O  the  line  ONJ_  to  the 
intersection  CD  of  the  two  planes.  Then  is  Z.  MON  the 
plane  angle  of  the  dihedral  A-CD-K.  (Why?) 


140 


Solid  Geometry. 


But  ZMON  is  a  right  angle  (Why?),  .'.  the  dihedral 
CD  is  a  right  dihedral,  and  hence  AB  is  J_  to  LK. 

Cor.  A  plane  perpendicular  to  the  edge  of  a  dihedral 
angle  is  perpendicular  to  both  its  faces. 

Proposition  XVIII. 

//  two  intersecting  planes  are  both  perpendicular  to  a 
third  plane,  their  intersection  is  perpendicular  to  that  plane. 

If  the  two  planes  CD  and  AB  intersecting  in  EF  are  both 
JL  to  GH,  then  EF  is  _[_  to  GH. 

Analysis:  If  we  can  identify  a  _L  to  GH  at  F  with  the 
intersection  of  CD  and  AB,  our  theorem  is  proved. 

Proof:   By   the  exercise   under   Prop.  XVI,   if   a  J_  be 


H 


Fig.  24- 

erected  to  GH  at  the  point  F  (common  to  both  AB  and 
CD)  it  will  lie  in  both  AB  and  CD,  that  is,  it  will  be  their 
intersection  and  identical  with  EF,  hence  EF  is  a  _[_  to  GH. 
Cor.  If  three  planes  are  mutually  perpendicular,  the 
intersection  of  any  two  is  perpendicular  to  the  third. 


Solid  Geometry. 
Proposition  XIX. 


141 


Every  point  in  a  plane  which  bisects  a  dihedral  angle  is 
equally  distant  from  its  faces. 

Statement:  Every  point  in  the  plane  AC  bisecting  the 
dihedral  angle  formed  by  the  planes  AN  and  AM,  is  equally 
distant  from  those  planes. 


M 


Fig.  25. 

Analysis:  This  proposition  suggests  a  Plane  Geometry 
proposition,  and  if  we  can  reduce  the  demonstration  to  a 
single  plane,  we  can  make  use  of  it. 

Proof:  Take  any  point  P  in  AC  and  draw  the  perpen- 
diculars PQ  and  PR  to  AN  and  AM  respectively.  Then 
we  are  to  prove  PQ  =  PR. 

Since  PQ  and  PR  intersect  at  P,  they  determine  a  plane 
DK,  which  is  _[_  to  both  AM  and  AN  (Why?),  hence  it 
is  perpendicular  to  their  intersection  AB,  or  AB  is  J_  to 
plane  DK. 

.-.  AB  is  J_  to  BR,  BQ  and  BP  (the  intersection  of  DK 
and  AC)  (Why?).  Hence,  Z  PBR  and  Z  PBQ  are 
the  plane  angles  respectively,  of  the  two  dihedrals  formed 
by  AC  and  AM,  and  by  AC  and  AN.  Therefore,  since 
these  two  dihedrals  are  equal,  being  halves  of  the  same 


142  Solid  Geometry. 

angle,  Z  PBR  =  Z  PBQ.  In  the  two  right  triangles 
PBQ  and  PBR,  Z  PBQ  =  Z  PBR,  and  PB  -  PB  (com- 
mon side). 

.*.  being  right  triangles  they  are  equal  throughout  and 
PQ  =  PR. 

Definitions. 

The  projection  of  a  -point  on  a  plane  is  the  point  where 
the  perpendicular  upon  that  plane  from  the  point  cuts  it. 

The  projection  of  a  line  on  a  plane  is  the  locus  of  the 
projection  of  all  its  points  on  that  plane.  The  pro- 
jection of  a  perpendicular  to  a  plane  on  that  plane  is  a 
point. 

The  projection  of  a  straight  line  that  does  not  cut  the 
plane,  is  a  straight  line  joining  the  feet  of  the  perpendicu- 
lars dropped  from  each  end  of  the  line  to  the  plane.  If 
the  line  cuts  the  plane,  its  projection  is  the  straight  line 
joining  its  point  of  intersection  in  the  plane,  with  the  foot 
of  the  perpendicular  dropped  from  the  end  (or  ends)  of  the 
line  upon  the  plane.  The  angle  which  a  line  makes  with 
a  plane  is  the  angle  it  makes  with  its  projection  on  the 
plane. 

Proposition  XX. 

Through  a  given  straight  line  not  perpendicular  to  a 
plane,  one  plane  can  be  passed  perpendicular  to  the  given 
plane. 

Statement:  If  AB  is  not  perpendicular  to  the  plane  EF, 
one  plane  can  be  passed  through  AB  J_  to  EF,  say  AD, 
intersecting  EF  in  CG. 

Analysis:  If  any  point  in  AB  be  selected  and  a  perpen- 
dicular drawn  from  it  to  the  plane,  any  plane  passed  through 
this  perpendicular  will  be  J_  to  the  plane  EF  by  a  previous 


Solid  Geometry. 


Fig.   26. 


proposition;  it  then  remains  to 
show  that  there  is  but  one  such 
plane. 

Prooj:  From  K,  any  point  in 
AB,  drop  the  perpendicular  KL 
to  the  plane  EF.  Through  this 
perpendicular  and  AB  pass  the 
F  plane  AD  intersecting  EF  in 
CG.  This  plane  will  be  J_  to 
EF.  It  is  the  only  plane  through 
AB  which  would  be  perpendicu- 
lar; because  if  two  planes  could  be  passed  through  AB  J_ 
to  EF,  AB  would  be,  of  course,  their  intersection  and 
would  be  J_  to  EF,  but  by  hypothesis  AB  is  not  _[_  to 
EF.  Hence  but  one  such  plane  can  be  drawn. 

Proposition  XXI. 

The  acute  angle  which  a  straight  line  makes  with  its  pro- 
jection  on  a  plane  is  the  least 
angle  it  makes  with  the  plane. 

Statement:  Let  BC  be  the  pro- 
jection of  AB  on  the  plane  MN, 
then  is  Z.  ABC  less  than  any 
other  angle  made  with  the  plane, 
say  Z  ABD. 

Analysis:  To  compare  angles 
or  lines,  whenever  possible,  it  is 
desirable  to  include  them  in  tri- 
angles whose  parts  may  be  compared. 

Proof:  Draw  BD  =  BC  and  AD   (AC  is  the  J_  deter- 
mining the  projection  of  AB  on  MN). 

In  the  two  triangles  ABC  and  ABD,  AB  is  common  and 


Fig.  27. 


144 


Solid  Geometry. 


BC  =  BD  (by  construction);  but  AC  <  AD  (Why?). 
Hence  Z  ABC  <  Z  ABD,  where  ABD  represents  any 
angle  made  by  AB  and  MN. 


Proposition  XXII. 


same  plane,  there 


Between  two  straight  lines  not  in 
can  be  only  one  common  perpendicular. 

Statement:  If  PR  and  CD  are  two  lines  not  in  the  same 
plane,  they  can  have  only  one  common  perpendicular. 
Analysis:  The  lines  must  be  compared  through  some 
p         x_  A    R  connecting   plane,  which    shall 

bear  a  known  relation  to  a 
plane  determined  by  one  of 
them,  and  it  is  clearly  desirable 
that  these  planes  be  J_. 

Proof:  Through  any  point  as 
rC    of    CD    draw  a    line    CE 
parallel  to  PR,  then  the  plane 
MN,  determined   by   CE   and 
Fig.  28.  CDj  is  parallel  to  PR  (Why?). 

Through  PR  pass  a  plane  PQ  perpendicular  to  MN  in- 
tersecting it  in  FG.  It  will  intersect  CD,  say  at  B,  other- 
wise CD  would  be  ||  to  PR,  contrary  to  hypothesis, 
which  says  they  are  not  in  the  same  plane.  At  B  erect  a 
_]_  AB  to  MN;  it  will  lie  in  PQ  by  a  previous  proposition. 
Now  AB  is  J_  to  CD  and  FG  (Why?),  but  FG  is  ||  to 
PR  (Why?),  hence  AB  is  _L  to  PR,  since  PR  and  FG  lie 
in  the  same  plane  and  are  ||.  Hence  AB  is  J_  to  both  PR 
and  CD.  If  it  is  not  the  only  perpendicular,  let  xC  (drawn 
from  any  other  point  x  in  PR  to  C)  be  J_  to  both  PR  and 
CD.  Then  *C  is_L  to  CD  and  CE  (Why?),  and  hence  J_ 


Solid  Geometry.  145 

to  MN.  From  x  draw  xy  J_  to  FG,  then  xy  is  _[_  to  MN 
(by  previous  proposition;  cite  it),  and  there  would  then 
be  two_J_'s  from  x  to  MN,  which  is  impossible. 

Definitions. 

The  space  included  by  three  or  more  planes  intersecting 
in  a  point  is  called  a  polyhedral  angle. 

The  point  of  intersection  is  called  its  'vertex  and  the 
planes  its  faces.  The  lines  in  which  adjacent  faces  inter- 
sect are  called  its  edges,  and  the  plane  angles  between  the 
edges  are  called  face  angles. 

The  size  of  the  polyhedral  angle  depends  solely  upon 
the  amount  of  divergence  of  its  faces  and  not  upon  their 
extent  nor  upon  their  shape. 

Each  pair  of  adjacent  faces  of  a  polyhedral  angle  form  a 
dihedral  angle. 

A  polyhedral  angle  is  convex  or  concave  according  as  a 
section  cutting  all  its  edges  is  convex  or  concave. 

A  polyhedral  angle  of  three  faces  is  called  a  trihedral 
angle;  of  four  faces,  a  tetrahedral  angle,  etc. 

A  trihedral  angle  may  evidently  have  one,  two,  or  three 
right  dihedral  angles. 

If  the  edges  of  a  polyhedral  angle  are  prolonged  through 
its  vertex,  another  polyhedral  angle  is  thus  formed  called  a 
symmetrical  polyhedral  angle,  with  its  faces  arranged  in 
reverse  order. 

Proposition  XXIII. 

The  sum  of  any  two  face  angles  of  a  trihedral  angle  is 
greater  than  the  third  face  angle. 

Statement:  In  the  trihedral  angle  A-KLM,  Z  KAL  + 
Z  LAM  >  Z  KAM. 

Analysis:  To  compare  these   angles  they  must  be  as 


146 


Solid  Geometry. 


usual  incorporated  into  triangles,  constructed  with  known 

parts  as  far  as  possible. 

Proof:  Let  KAM  be  the  greatest  of  all  the  angles,  and  in 

the  plane  of  KAM  draw 
AN,  making  Z  KAN  = 
ZKAL.  Lay  off  AD  (on 
AN)  =  AC  (on  AL)  any  con- 
venient distance.  Through 
D,  draw  any  line  BDE 
cutting  AK  at  B  and  AM 
at  E.  Through  BDE  and 
C  pass  a  plane,  intersecting 
^face  KAL  in  BC  and  LAM 
in  CE.  Then  in  the  trian- 
gles BAG  and  BAD, 

AB  =  AB  (common). 
AC  =  AD  (by  construction). 
Z  BAG  =  Z  BAD  (by  construction). 
.-.  A  BAG  =  A  BAD,  hence  BC  =  BD. 
But     BC  +  CE  >  BD  +  DE  (or  BDE). 
Subtract,        BC  =  BD. 
CE  >  DE. 
In  the  two  triangles  CAE  and  DAE, 

AC  =  AD  (by  construction) 
AE  =  AE  (common). 
But  CE  >  DE  (by  above). 

.'.   Z  CAE  >  Z  DAE.  (Why?) 

Add         Z  BAG  =  Z  BAD. 

Z  BAG  +  Z  CAE  >  Z  BAD  +  DAE 
or        Z  KAL  +  Z  LAN  >  KAM. 


Solid  Geometry.  147 

Proposition  XXIV. 

The  sum  of  all  the  face  angles  of  any  polyhedral  angle  is 
less  than  jour  right  angles. 

Statement:  If  A  is  a  polyhedral  angle,  then  BAG  +  CAD 
-I-  DAE  +  EAF  +  FAG  +  GAB  is  <  four  right  angles, 
B,  C,  D,  E,  F  and  G  being  any  points 
on  its  edges. 

Analysis:  To  estimate  parts  of  a 
polyhedral  angle,  the  parts  must  be 
definitely  enclosed  and  defined,  and  that 
can  be  done  only  by  passing  a  plane 
through  the  polyhedral  cutting  all  edges. 

Proof:  Pass  a  plane  through  the 
points  B,  C,  D,  E,  F  and  G,  thus 
forming  the  polygon  BCDEFG.  Take 
any  point  K  within  this  polygon  and  Fig.  30. 

connect  it  with  the   vertices  B,  C,  D,  E,  F  and  G,   thus 
forming  triangles  with  the  common  vertex  K. 

At  each  vertex,  B,  C,  D,  etc.,  are  trihedral  angles,  where 
by  the  last  proposition,  the  sum  of  two  face  angles  is  greater 
than  the  third.  Hence, 

Z  BCA  +  Z  ACD  >  Z  BCD 
or  Z  BCA  +  Z  ACD  >  Z  BCK  +  Z  KCD. 

Z  ADC  +  Z  ADE  >  Z  CDE 
or  Z  ADC  +  Z  ADE  >  Z  KDC  +  Z  KDE. 

Z  AED  +  Z  AEF  >  Z  DEF,  etc. 

That  is,  the  sum  of  all  the  angles  at  the  base  of  the  tri- 
angles having  vertex  A  is  greater  than  the  sum  of  all  the 
angles  at  the  base  of  the  triangles  having  vertex  K. 

But  the  sum  of  all  the  angles  of  the  triangles  with  vertex 
A  is  equal  to  the  sum  of  all  the  angles  of  triangles  with 


148 


Solid  Geometry. 


vertex  K,  because  there  are  the  same  number  of  triangles 
in  each  and  the  sum  of  the  angles  of  any  triangle  equals 
two  right  angles,  hence  the  sum  of  the  angles  at  A  is  less 
than  the  sum  of  the  angles  at  K.  This  can  be  represented 
thus :  Call  the  sum  of  the  base  angles  of  triangles  having 
vertex  K,  X;  and  the  sum  of  the  base  angles  of  triangles 
having  vertex  A,  Y;  then  X  <  Y  (by  what  has  been  proved). 
Call  the  sum  of  all  the  angles  of  triangles  having  vertices 
K  and  A,  each  Z  (for  it  is  the  same  in  both). 


Then 


Z=  Z 
X<  Y. 


Subtract;    Z  -  X  >  Z  -  Y  (the  less  subtrahend  leaves 
greater  remainder). 

But  Z  —  X  =  sum  of  angles  at  K 

and  Z  —  Y  =  sum  of  angles  at  A. 

The  sum  of  the  angles  at  K  =  4  right  angles.         (Why?) 

.*.  The  sum  of  the  angles  at  A  <  4  right  angles. 

Proposition  XXV. 

Two  trihedral  angles  are  equal  if  the  three  /ace  angles  of 
one  are  equal  to  the  three  face  angles  of  the  other. 


N 
Fig-  31. 

Statement:  If  in  the  two  trihedrals  A  and  B,  /_  CAD 


Solid  Geometry.  149 

Z  MEN,     Z  DAE  =  Z  NBO,     and  Z  CAE  =  Z  MBO, 

then  Z  A  =  Z  B. 

Analysis:  The  two  trihedrals  are  evidently  equal  if  their 
dihedral  angles  are  equal,  for  then  they  will  exactly  coin- 
cide, when  superposed;  and  the  dihedrals  are  equal  if  their 
plane  angles  are  equal,  hence  we  must  make  such  a  con- 
struction that  the  plane  angles  of  two  corresponding  dihe- 
drals can  be  compared,  and  that  requires  equal  triangles 
including  those  angles.  We  are  clearly  at  liberty  to  con- 
struct these  as  simply  as  possible. 

Proof:  Lay  off  from  A  and  B  equal  distances  on  all  the 
edges,  say  AC,  AD,  AE,  BM,  BN  and  BO.  Join  C,  D 
and  E,  also  M,  N  and  O. 

Then  the  triangles  CAD  and  MEN,  DAE  and  NBO 
and  CAE  and  MBO  have  each  two  sides  and  the  included 
angle  equal,  hence  are  equal. 

Hence   CD  -  MN,  DE  =  NO  and  CE  =  MO. 

Also      Z  ACD  -  Z  BMN,  Z  ACE  =  Z  BMO,  etc. 

Then  triangles  CDE  and  MNO  have  three  sides  in 
each  equal,  hence  they  are  equal  and  Z  ECD  =  Z  OMN. 

On  AC  and  BM,  respectively,  lay  off  CF  =  MP,  and  at 
F  and  P  construct  the  plane  angles  of  the  dihedrals  AC 
and  BM,  HFG  and  RPQ.  The  sides  of  HFG  intersect 
CE  in  G  and  CD  in  H,  and  the  sides  of  RPQ  intersect 
MO  in  Q  and  MN  in  R.  (Why?) 

Join  G  and  H,  and  R  and  Q  by  straight  lines. 

In  the  right  triangles,  HCF  and  RMP, 

CF  =  MP  (by  construction). 
Z  FCH  =  Z  PMR  (by  above  proof). 

/.  triangle  HCF  =  triangle  RMP,  and  hence  FH  =  PR, 
CH  =  MR. 

The  right  triangles  CFG  and  MPQ  are  equal  for  same 
reason,  hence  FG  =  PQ  and  CG  =  MQ. 


150  Solid  Geometry. 

In  the  triangles  (not  right)  HCG  and  RMQ 
CH  =  MR  (by  above  proof). 
CG  =  MQ  (by  above  proof). 
Z  HCG  =  Z  RMQ  (by  proof). 
/.    triangle  CHG  =  triangle  RMQ  and  HG  =  RQ. 

(Why?) 

In  triangles  HFG  and   RPQ,  FH  =  PR, 
FG  =  PQ,    HG  =  RQ. 
.*.  triangle  HFG  =  triangle  RPQ 

and  Z  HFG  =  Z  RPQ.  (Why  ?) 

/.  dihedral  angle  AC  =  dihedral  BM. 

In  the  same  way  all  the  other  dihedrals  may  be  proved 
equal.  It  will  be  observed  how  all  the  demonstration  is 
concentrated  upon  the  proving  of  equality  between  the 
parts  of  triangles  HFG  and  RPQ,  and  how  the  construc- 
tion was  made  to  assist  in  this  aim. 

EXERCISE    II. 

Prove: 

1.  The  planes  that  bisect  two  supplementary-adjacent 
dihedral  angles,  are  perpendicular  to  each  other. 

2.  Show  that  a  plane  perpendicular  to  the  edge  of  a 
dihedral  angle  is  perpendicular  to  both  faces. 

3.  Prove  the  converse  of  2. 

4.  What  is  the  greatest  number  of  equilateral  triangles, 
squares  and  pentagons,  respectively,  that  can  be  grouped 
about  a  point  to  form  a  polyhedral  angle. 

5.  From  two  given  points  on  the  same  side  of  a  given 
plane,  draw  two  lines  to  a  point  in  the  plane,  meeting  it 
at  equal  angles.     What  is  the  locus  of  the  point  of  meeting 
in  the  plane? 

6.  What  is  the  locus  of  points  in  space  equally  distant 
from  two  intersecting  lines  ? 


Solid  Geometry.  151 

7.  Find  a  point  equally  distant  from  four  given  points 
not  in  the  same  plane. 

8.  If  a  line  is  parallel  to  one  plane  and  perpendicular 
to  another,  the  two  planes  are  perpendicular  to  each  other. 

9.  Find   the   locus  of  points  in    space  equally  distant 
from  two  planes  (two  cases). 

10.  Find  the  locus  of  points  equally  distant  from  two 
lines  not  in  the  same  plane. 

11.  Show  that  a  plane  may  be  drawn  perpendicular  to 
only  one  edge,  and  to  only  two  faces  of  a  polyhedral  angle. 

12.  If  a  straight  line  is  parallel  to  a  plane,  any  plane 
perpendicular  to  the  line  is  perpendicular  to  the  plane. 

13.  If  a  line  is  inclined  to  a  plane  at  an  angle  of  60°, 
its  projection  on  the  plane  is  equal  to  half  the  line. 


GEOMETRICAL    SOLIDS. 

Definitions. 

A  polyhedron  is  a  solid  bounded  by  planes. 

A  diagonal  is  a  straight  line  joining  any  two  vertices  of 
a  polyhedron  not  in  the  same  face. 

A  section  of  a  polyhedron  is  the  portion  of  an  intersect- 
ing plane  inclosed  by  its  intersections  with  the  faces  of  the 
polyhedron. 

A  polyhedron  is  convex  if  every  section  of  it  is  a  convex 
polygon. 

A  polyhedron  of  four  faces  is  called  a  tetrahedron.  A 
polyhedron  of  five  faces  is  called  a  pentahedron.  A  poly- 
hedron of  twenty  faces  is  called  an  icosahedron,  etc. 


152  Solid  Geometry. 

PRISMS   AND    PARALLELOPIPEDS. 

A  prism  is  a  polyhedron,  two  faces  of  which  are  equal 
polygons  in  parallel  planes,  and  the  other  faces  are  parallel- 
ograms. 

The  equal  polygons  are  usually  called  the  bases;  the 
parallelograms,  lateral  faces;  and  the  intersections  of  these 
faces,  lateral  edges. 

The  sum  of  the  lateral  faces  is  the  lateral  area. 

A  right  prism  is  one  whose  lateral  edges  are  perpendicu- 
lar to  the  planes  of  the  bases.  A  prism  whose  edges  are 
not  perpendicular  to  the  bases  is  an  oblique  prism. 

A  regular  prism  is  a  right  prism  with  regular  bases. 
Prisms  are  named  by  their  bases,  triangular,  quadrangular, 
etc. 

A  parallelepiped  is  a  prism,  whose  bases  are  parallelo- 
grams. A  right  parallelepiped  is  one  whose  lateral  edges 
are  perpendicular  to  the  planes  of  its  bases;  if  its  6  faces 
are  all  rectangles,  it  is  called  a  rectangular  parallelepiped. 

A  cube  is  a  parallelepiped  whose  faces  are  squares.  A 
cube  whose  edges  are  a  linear  unit  in  length  has  been 
chosen  as  the  unit  of  'volume. 

The  altitude  of  any  prism  is  the  perpendicular  distance 
between  its  bases. 

The  volume  of  any  solid  is  the  number  of  units  of  volume 
which  it  contains. 

Two  solids  having  equal  volumes,  are  said  to  be  equi- 
valent (not  equal).  When  would  they  be  equal? 

A  right  section  of  a  prism  is  a  section  perpendicular 
to  the  lateral  edges. 

A  truncated  prism  is  a  portion  of  a  prism  included 
between  its  base  and  any  oblique  section. 


Solid  Geometry . 


SCHEMATIC    ARRANGEMENT. 


Polyhedron 


Prism 


Right 


Regular 
Irregular 


Oblique 
Parallelepiped 


JRight 
[Oblique 


( Rectangular — Cube 
( Parallelogram — Bases 


Proposition  XXVI. 

The  sections  of  a  prism  made  by  parallel  planes  cut- 
ting all  the  lateral  edges  are  equal. 

Statement:  In  the  prism  AB,  if  the 
sections  12345  and  67890  are  ||  they 
are  equal. 

Analysis:  Equal  figures  must  be 
capable  of  coincidence  if  superposed. 
Hence  must  have  all  parts  equal. 

Proof:  The  sides  12,  23,  34,  45  and 
51  are  ||  respectively  to  67,  78,  89,  90 
and  06  (Why?),  and  since  the  edges 
of  the  prism  are  parallel  by  definition 
12,  23,  34,  etc.,  are  equal  to  67,  78, 
89,  etc.,  respectively.  The  angles 
123,  234,  345,  451  and  512  are  re- 
spectively equal  to  678,  789,  890,  906, 
067  (Why?).  Hence  the  two  poly- 
gons having  equal  sides  and  angles 
could  be  made  to  coincide. 

Cor.  Every  section  of  a  prism  parallel  to  the  bases  is 
equal  to  the  bases;  and  all  right  sections  of  a  prism  are 
equal,. 


Fig.  32. 


154 


Solid  Geometry. 


Fig.  33- 


Proposition   XXVI  (a). 

The  lateral  area  of  a  prism  is  equal  to  the  -perimeter  of  a 

right  section  multiplied  by 
a  lateral  edge. 

Statement:  If  DEFGH 
is  a  right  section  of  the 
prism  AC,  then  the  lateral 
area  =  (DE  +  EF  +  FG 
+  GH  +  HD)  X  BK  (or 
any  other  edge). 

Analysis:  Since  the  lat- 
eral area  is  made  up  of 
parallelograms,  and  the 
area  of  a  parallelogram  is 
equal  to  the  product  of  base  and  altitude,  it  is  simply  a 
question  of  determining  the  bases  and  altitudes  of  these 
component  parallelograms. 

Proof:  By  definition  of  right  section,  the  sides  of  the 
section  are  all  perpendicular  to  the  edges,  hence  taking 
the  edges  as  bases  of  the  parallelograms  these  sides  will 
be  their  altitudes. 

Hence  face  AB  =   KB  X  DH, 

"    KC  -   KB  X  HG, 

"    LP  -.    LC  X  GF, 

"   OR  -    OP  X  FE, 

"  MN  =  MR  X  ED, 

and  so  on  for  any  number  of  faces.     Adding;  lateral  area  = 
KB  X  DH  +  KB  X  HG  +  LC  X  GF  +  OP  X  FE  + 
MR  x  ED  =  (DH  +  HG  +  GF  +  FE  +  ED)  X  KB, 
since  all  the  edges  are  equal.  (Why?) 


Solid  Geometry. 


Proposition  XXVII. 

Any 'oblique  prism  is  equivalent  to  a  right  prism  whose 
base  is  equal  to  a  right  section  of  the  oblique  prism,  and 
whose  edges  (or  altitude)  are  equal 
to  the  edges  of  the  oblique  prism. 

Statement:  If  the  prism  i  — 
67890  is  a  right  prism  having  a 
base  67890  =  a  right  section  of  the 
prism  A  —  MNRPQ  and  an  alti- 
tude (same  as  an  edge  in  a  right 
prism)  equal  to  an  edge  of  the 
prism  A  —  MNRPQ,  then  are 
these  prisms  equivalent. 

Analysis:  The  usual  test  of 
geometrical  values,  superposition 
for  comparison.  As  in  the  nature 
of  the  case  exact  coincidence  is 
impossible,  the  next  best  thing  is 
to  bring  the  figures  together  in 
order  to  compare  parts. 

Proof :  Combine  the  two  prisms 
so  that  a  base  of  the  right  prism 


Fig.  34- 


will  become  a  right  section  of  the  oblique  prism;  then  the 
edges  of  the  two  prisms  will  take  the  same  direction 
(Why?),  the  edges  of  the  right  prism  extending  beyond  the 
upper  base  of  the  oblique  prism,  so  that 


Ai  =  M6,  B2  =  Ny,  C3  =  R8,  etc. 


(Why?) 


The  entire  solid  is  made  up  of  a  portion  of  each  original 
prism  and  a  common  truncated  prism  A  —  67890.  An 
examination  of  the  figure  will  show  that  if  A  —  67890  is 
added  to  6  —  MNRPQ,  the  result  is  the  oblique  prism; 


156  Solid  Geometry. 

and  if  it  is  added  to  i  -  ABCDE,  the  result  is  the  right 
prism,  hence  if  it  can  be  shown  that  6  —  MNRPQ  = 
i  —  ABCDE,  our  proposition  is  easily  established. 

As  always,  superposition  will  make  it  possible  to  com- 
pare these  two  truncated  prisms  6  —  MNRPQ  and  i  - 
ABCDE.  Place  the  base  MNRPQ  of  one  upon  ABCDE 
the  base  of  the  other,  Q  on  E;  they  will  exactly  coincide, 
for  they  are  the  bases  of  the  original  oblique  prism, 
and  hence  are  equal.  The  lateral  faces,  MQo6,  QPpo, 
PR89,  etc.,  are  equal  respectively  to  the  faces,  AE5i, 
ED45,  DC34,  etc.  (Why?) 

Hence  the  edges  M6,  Qo,  PQ,  etc.,  will  fall  upon  the 
edges  Ai,  £5,  D4,  etc.  (since  the  face  angles  of  the 
trihedrals  A,  E,  D,  etc.,  are  equal,  these  trihedrals  are 
equal  by  former  Proposition).  Since  M6  =  Ai,  Qo  =  £5, 
Pp  =  D4,  etc.,  6  will  fall  on  i,  o  on  5,  9  on  4,  etc.,  hence 
the  bases  67890  and  12345  must  coincide, 

.  6  -  MNRPQ  =  i   -  ABCDE; 
and  A  —         67890  =  A  —    67890 

adding  ;  A  -  MNRPQ  =  i   -      67890. 

Cor.  I.  Two  prisms  having  the  three  faces  of  a  tri- 
hedral angle  in  one  equal  to  the  three  faces  (similarly 
placed)  of  a  trihedral  angle  in  the  other  are  equal. 

Cor.  II.  Two  right  prisms  are  equal  if  they  have  equal 
bases  and  altitudes. 

EXERCISE. 

Show  that  the  opposite  faces  of  a  parallelepiped  are 
equal  and  parallel. 


Solid  Geometry. 


Proposition   XXVIII. 

//  a  plane  be  passed  through  two  diagonally  opposite 
edges  of  a  parallelepiped,  it  will  divide  the  parallelepiped 
into  two  equivalent  triangular  prisms. 

Statement:  Let  the  plane  ED  be  passed  through  the 
edges  BE  and  DG  of  the  parallelepiped  B  —  EFGH, 
then  the  two  triangular  prisms  so  formed,  C  —  EGH  and 
A  —  EFG  are  equivalent. 

Analysis:  Prop.  XXVII  and 
Cor.  II  under  it,  give  us  a  clue  to 
the  comparison  of  prisms  in 
general,  through  the  medium  of 
right  sections,  which  act  as  bases 
for  right  prisms. 

Proof:  Since  it  is  necessary  to 
express  oblique  prisms  in  terms 
of  right  prisms,  in  order  to  com- 
pare them,  Prop.  XXVII  suggests 
a  method. 

Make  a  right  section,  1234 
of  the  parallelepiped;  it  will 
cut  the  plane  ED  in  its  diagonal 

13,  which  will  divide  it  into  equivalent  triangles,  since  it 
will  be  a  parallelogram.  (Why?)  Then  these  triangles 
will  be  right  sections  of  the  two  triangular  prisms. 

The  triangular  prism  C  —  EHG  is  equivalent  to  a 
right  prism,  whose  base  is  the  right  section  143,  and  whose 
altitude  is  an  edge,  and  A  —  EFG  is  equivalent  to  right 
prism  with  base  123.  But  143  =  123,  and  the  edges  are  all 
equal,  .'.  these  two  right  prisms  will  be  equivalent,  hence 
the  prisms,  C  —  EHG  and  A  —  EFG  are  equivalent. 


pig.  35. 


158 


Solid  Geometry. 


Proposition  XXIX. 

Two  rectangular  parallelepipeds  having  equal  bases  are 
to  each  other  as  their  altitudes. 

Statement:  If  the  bases  of  the  two  parallelepipeds,  D  - 
EFGH  and  M  -  QTSR  are  equal,  then  D  -  EFGH  : 
M  -  QTSR  :  :  DE  :  MQ. 


Fig.  36. 

Analysis:  There  are  plainly  two  cases:  when  the  altitudes 
(edges)  are  commensurable  and  when  they  are  incom- 
mensurable. In  the  first  case  it  is  necessary  only  to  apply 
the  common  measure  and  in  the  second  case  to  take  the 
unit  of  measure  so  small  that  we  may  approach  commen- 
surability  as  near  as  we  please. 

Proof:  Case  I: 

Apply  the  common  measure  to  the  altitudes  (edges) 
DE  and  MQ;  say  it  is  contained  in  DE,  x  times,  and  in 
MQ,  y  times,  then, 

DE  :  MQ  \\x\y. 

Through  the  points  of  division  pass  planes  ||  to  the  bases; 
they  will  divide  the  parallelepipeds  into  x  and  y  parts, 
respectively,  which  will  all  be  equal.  (Why?) 


Solid  Geometry, 


159 


Hence,        D  -  HEFG:  M  -  RQTS:  :  oc:  y. 
and  D  -  HEFG:  M  -  RQTS:  :  DE:  MQ. 

Case  II  is  left  to  the  student. 

Cor.  Since  the  area  of  a  rectangle  equals  the  product 
of  its  two  dimensions,  Prop.  XXIX  may  be  stated  thus: 
Two  parallelepipeds,  having  two  dimensions  equal,  are 
to  each  other  as  the  third  dimension. 


Proposition    XXX. 

Two  rectangular  parallelepipeds  having  equal  altitudes  are 
to  each  other  as  their  bases. 


Fig.  37. 

Statement:  Let  the  parallelepipeds  A  and  B   have  the 
same  altitude  m  and  bases  y  X  z  and  /  X  2'. 


160  Solid  Geometry. 

Analysis:  Since  any  face  of  a  parallelepiped  may  be 
taken  as  its  base,  another  parallelepiped  can  be  constructed 
having  two  dimensions  in  common  with  both  A  and  B, 
with  which  both  may  be  compared. 

Proof:  Construct  the  rectangular  parallelepiped  C  with 
the  dimensions  m,  yf,  z. 

Then  taking  m  X  2  as  the  base  of  C,  —  =  ^  (Prop.  29)  ; 

C       z 

and  taking  /  X  m  as  base  of  C,  —  =  —  ,  (by  Prop.  29). 

B       z 

By  multiplying  these  two  equations  together, 
A  v£  _y  vz      .A  _y  X  z 
X~X       (          ~' 


Cor.  Prop.  XXX  may  evidently  be  stated  thus: 
Two  rectangular  parallelepipeds  having  one  dimension 
equal  are  to  each  other  as  the  product  of  the  other  two. 


Proposition  XXXI. 

Two  rectangular  parallelepipeds  are  to  each  other  as  the 
product  oj  their  three  dimensions. 

Statement:  If  the  two  parallelepipeds,   M  and  N,  have 
the  dimensions  a,  &,  c  and  a',  b',  c',  respectively,  then 
M    =  a  X  b  X  c 
N    :~  a'  X  V  Xc'  ' 

Analysis:    Compare    each    with    another    parallelepiped 
having  partly  the  dimensions  of  both. 

Proof:  Construct  the  parallelepiped  R  with  the  dimen- 
sions a,  b,  and  c'.     Then, 

=  i  and    R= 


Re'  N       a'  X  A' 


Solid  Geometry. 


161 


multiplying  these  equations, 


X 


or 


M. 

N 


a   X  b   X 


N       a'XVXc'          N       a'XVXc' 
Cor.     The  volume  of  a  rectangular  parallelepiped  equals 
the  product  of  its  three  dimensions.     For  suppose  a  certain 


Fig.  38. 

parallelepiped  A  has  the  dimensions  a,  b,  c,  and  a  cube  B 
has  the  edges  each  one  unit  in  length  (the  same  unit  in 
which  a,  b  and  c  are  expressed)  or  has  the  dimensions  i, 
1,1.  By  Prop.  XXXI, 

A  =  a  X  b  X  c 

B 


=   a  X  b  X  c. 


i  X  i  X  i 
But  volume  is  defined  as  the  number  of  times  a  solid 

contains  the  unit  of  volume,  which  is  universally  taken  to 

^ 
be  a  cube,  whose  edges  are  one  unit  in  length,  hence  —  = 

the  volume  of  A  =  a  X  b  X  c. 


162 


Solid  Geometry. 


Proposition  XXXII. 

The  volume  of  any  parallelepiped  is  equal  to  the  product 
of  its  base  and  altitude. 

Statement:    The    oblique    parallelepiped    D  -  EGFH  = 
EGFH  X  OP,  OP  being  its  altitude,  as  indicated  in  Fig.  39. 


Fig.  39- 

Analysis:  Since  it  is  known  that  the  volume  of  a  rec- 
tangular parallelepiped  equals  the  product  of  its  three 
dimensions,  which  is  equivalent  to  saying  the  product  of 
its  base  by  its  altitude,  it  is  necessary  to  express  the  oblique 
parallelepiped  in  terms  of  an  equivalent  rectangular  paral- 
lelepiped. Also  it  is  known  that  an  oblique  parallelepiped 
is  equivalent  to  a  right  parallelepiped  whose  base  equals  a 


Solid  Geometry.  163 

right  section  of  the  oblique  and  whose  altitude  is  an  edge. 
If  right  sections  be  taken  successively  perpendicular  to 
each  other  in  the  oblique  parallelepiped,  the  result  will 
clearly  be  a  rectangular  parallelepiped,  no  matter  how 
oblique  the  original  may  have  been.  It  remains  only  to 
prove  the  equivalence  of  these  successive  figures. 

Proof:  Through  the  vertex  B  pass  the  right  section  BJLK 
and  move  it  out  parallel  to  its  first  position  along  the 
produced  edges  BA,  CD,  FH  and  GE  to  x ;  then  on  these 
produced  edges  lay  off  KM  =  FH,  BN  =  BA,  LQ  =  GE 
and  JR  =  CD  and  draw  the  equivalent  right  section 
MNRQ.  B  -  LQMK  is  then  a  right  parallelepiped 
equivalent  to  D  —  EGFH,  for  its  base  (any  face  of  a 
parallelepiped  may  be  its  ibase)  BJLK  is  a  right  section 
of  D  -  EGFH  and  its  altitude  BN  is  equal  to  the  edge 

CD. 

B  —  LQMK  is  not  yet  rectangular. 

Through  the  edge  JR  pass  the  right  section  JR  21  and 
move  it  out  to  y,  along  the  produced  edges.  On  these 
produced  edges  lay  off  R  6,  J  5,  14,  and  23  equal  to  the 
edges  BJ,  etc.,  and  draw  the  equivalent  right  section 
3456  forming  the  rectangular  parallelopiped  J  —  1234. 

Now  these  three  parallelepipeds  all  have  the  same  alti- 
tude OP,  because  the  planes  have  remained  unchanged 
in  direction;  also  the  bases  1234,  KMQL  and  EGFH 
are  all  equal,  because  the  sides  23  and  MQ  and  LQ  and 
GE  are  equal  by  construction,  and  the  opposite  sides  lie 
in  parallel  lines,  hence  their  altitudes  are  the  same. 

But        J  -  1234  =  base  X  alt.  =  1234  X  OP, 
and  since  the  parallelepipeds  are  all  equivalent, 

D  -  EGFH  =  J  -  1234  =  1234  X  OP  = 

EGFH  X  OP; 
that  is,        D  -  EGFH  =  EGFH  X  OP. 


164 


Solid  Geometry. 


Proposition  XXXIII. 

The  volume  of  a  triangular  prism  is  equal  to  the  product 
of  its  base  by  its  altitude. 

Statement:  The  volume  of  the  triangular  prism  C  —  FGH 
=  FGH  X  OP. 

Analysis:  It  is  clearly  necessary  to  express  the  triangular 
prism  in  terms  of  a  parallelepiped,  since  we   have  deter- 
mined the  volume  of  the  latter.     The 
relation    between   a   triangle    and  a 
parallelogram  of  the  same  base  and 
altitude,  will  suggest    a   method   of 
c  converting  the  triangular  prism  into 
a  parallelopiped. 

Proof:  Complete  the  parallelograms 
of  which  the  triangular  bases  of  the 
prisms  are  halves. 

Thus,  draw  AB  ||  to  DC;  AD  ||  to 
BC;  EF  ||  to  HG  and  EH  ||  to  FG  in 
the  planes  of  the  triangles  BDC  and  FGH. 

Join  A  to  E,  and  we  have  a  parallelopiped  A  —  EFGH, 
whose  bases  are  twice  the  bases  of  the  prism  (Why?),  and 
which  has  the  same  altitude,  OP. 

The  volume  of  this  parallelopiped  =  EFGH  X  OP. 
But      EFGH  =  2  FGH  and  by  Prop.  XXVIII 
Vol.  A  -  EFGH  =  2  vol.  C  -  FGH. 

/.  Vol.  A  -  EFGH  =  EFGH  X  OP  can  be  expressed 
thus: 

2  Vol.   C  -  FGH  =  2  FGH  X  OP. 
/.  Vol.  C  -  FGH  =  FGH  X  OP. 


Fig.  40. 


Solid  Geometry. 


165 


Proposition  XXXIV. 

The  volume  of  any  prism  equals  the  product  of  base  by 
altitude. 

Statement:  The  volume  of  the  prism  E  —  FGHKL  = 
the  base  FGHKL  multiplied  by  the  altitude  M. 

Analysis:  Since  every  prism  may 
be  divided  into  triangular  prisms 
by  passing  planes  through  the 
non-adjacent  edges,  the  character- 
istics of  triangular  prisms  belong 
to  any  prism. 

Proof:  Through  the  edges  EL 
and  CH,  also  EL  and  BG,  pass  M 
planes  dividing  the  prism  into  the 
triangular  prisms  A  —  FGL,  E  — 
GLH  and  D  -  LKH,  having  the 
same  altitude  M,  as  the  original^ 
prism,  and  the  bases  FGL,  GLH 
and  LKH  make  up  the  base, 
FLKHG,  of  the  original  prism. 
Also  the  original  prism  equals  the 
sum  of  the  triangular  prisms.  Fig.  41. 

Now    A  -  FLG  =  FLG  X  M;  E  -GLH  = 

GLH  X  M  and  D  -  LKH  =  LKH  X  M 
and  E  -  FLKHG  =  A  -  FLG  +  E  -  GLH  +D-LKH, 
also  FGHKL  =  FGL  +  GLH  +  LKH. 

Hence,  E  -  FLKHG  -  (FLG  +  GLH  +  LKH)X  M 
or  E  -  FLKHG  =  FGHKL  XM. 

Definitions. 

A  pyramid  is  a  polyhedron  whose  base  is  a  polygon 
and  whose  lateral  faces  are  triangles  having  a  common 
vertex.  Draw  one.  The  intersections  of  the  faces  are 


i66 


Solid  Geometry. 


called  edges,  as  usual;  and  the  common  vertex  is  known 
as  the  apex,  or  the  vertex,  of  the  pyramid.  The  sum  of 
the  lateral  faces  is  called  the  lateral  area. 

The  altitude  of  a  pyramid  is  the  perpendicular  distance 
from  the  apex  to  the  plane  of  the  base. 

Pyramids  are  named  according  to  their  bases;  triangular, 
quadrangular,  pentagonal,  etc. 

A  triangular  pyramid  is  also  called  a  tetrahedron,  since 
it  has  four  faces. 

A  pyramid  whose  base  is  a  regular  polygon  with  its 
centre  at  the  foot  of  the  perpendicular  from  the  apex  to 
the  base,  is  called  a  regular  pyramid.  This  perpendicular 
from  the  apex  to  the  base-centre  is  called  the  axis  of  the 
regular  pyramid. 

The  lateral  faces  of  a  regular  pyramid  are  equal  isosceles 
triangles  (Why?). 

The  slant  height  of  a  regular  pyramid  is  the  altitude  of 
any  one  of  these  isosceles  triangles. 
Would   an  oblique   pyramid   have 
a  slant  height  ? 

A  truncated  pyramid  is  the  portion 
of  a  pyramid  included  between  the 
base  and  any  section  made  by  a 
plane  cutting  all  lateral  edges.  When 
the  section  is  parallel  to  the  base, 
the  truncated  pyramid  is  called  a 
^)cfrustrum  of  a  pyramid. 

Proposition  XXXV. 


E  D  The  lateral  area  oj  a  regular  pyra- 

mid is  equal  to  half  the  product  of  the 
slant  height  and  the  perimeter  of  the  base. 


Solid  Geometry. 


167 


Statement:  If  A  —  BCDEF  is  a  regular  pyramid,  whose 
slant  height  is  AG  then  its  lateral  area  is 

i  (BCDEF  X  AG). 

Analysis:  By  definition  of  regular  pyramid  and  slant 
height,  the  relation  between  the  slant  height  and  the  tri- 
angular faces  is  evident. 

Proof:  Each  face  being  an  isosceles  triangle,  all  having 
the  same  altitude;  namely,  the  slant  height  of  the  pyramid, 
its  area  will  equal  J  its  base  X  slant  height,  and  its  base 
will  be  one  side  of  the  base  of  the  pyramid.     That  is, 
ABC  =  iBC  X  AG; 
ACD  =  J  CD  X  AG; 
ADE  -  J  DE  X  AG;  etc. 

Adding;  lateralarea=  J  (BC  +  CD  +  DE  +  EF  +  FB)  X  AG. 
Cor.     The   lateral   area   of   the   frustrum   of   a   regular 
pyramid  equals  one-half  the  sum  of  the  perimeters  of  the 
bases  multiplied  by  the  slant  height. 

What  is  the  shape  of  the  faces  of  the  frustrum  ? 


Proposition   XXXVI. 

A  section  parallel  to  the  base  of  a 
pyramid  is  similar  to  the  base  and 
divides  the  edges  and  altitude  pro- 
portionally. 

Statement:  Let  MNOPQ  be  a  sec- 
tion of    the  pyramid    A  -  BCDEF 
|!   to   BCDEF   cutting    the     altitude 
at  R.     Then  AM  :  AB  :  :  AN  :  ACF 
:  :  AO  :  AD  :  :  AR  :  AT,  etc. 

Analysis :  Parallel  planes  cut  by 
transversal  planes  suggest  the  rela- 
tions of  the  lines  of  intersection  in 
the  triangular  faces,  and  hence  the 


Fig.  43- 


1 68  Solid  Geometry. 

relations  between  the  section-polygon  and  the   base-poly- 
gon. (Why?) 
Proof: 
Since       MN  is  ||  to  BC,  NO  ||  to  CD,  OP  ||  to  DE,  etc. 

AM  :  AB   :  :  AN  :  AC 
Also  AN  :  AC  :  :  AO  :  AD 

AO  :  AD  :  :  AP  :  AE,  etc. 
Hence, 

AM  :  AB  :  :  AN  :  AC  :  :  AO  :  AD  :  :  AP  :  AE,  etc., 
and  in  the  right  A's  AMR  and  ABT, 

AM  :  AB  :  :  AR  :  AT.          (Why?) 

2dPart: 

Since  MN  is  ||  to  BC,  NO  ||  to  CD,  PO  ||  to  ED,  etc. 

Z  MNO  =  Z  BCD,  Z  NOP  =  Z  CDE,  etc.       (Why  ?) 

Also  MN  :  BC  :  :  AN  :  AC 

and  NO  :  CD  :  :  AN  :  AC,  etc. 

/.  MN  :  BC  :  :  NO  :  CD,  etc. 

Hence,  MNOPQ  and  BCDEF  are  equiangular  and 
have  proportional  sides,  .'.  are  similar. 

Cor.  I.  Any  section  of  a  pyramid  parallel  to  the  base, 
is  to  the  base  as  the  square  of  its  distance  from  the  vertex 
is  to  the  square  of  the  altitude. 

That  is,      MNOPQ  :  BCDEF  :  :  AR2  :  AT2. 

For  MNOPQ  and  BCDEF  being  similar  polygons, 

MN  :  BC  :  :  AM  :  AB  :  :  AR  :  AT 
or  MN2:BC2::AR2:^T2._ 

.'.  MNOPQ  :  BCDEF  :  :  AR2  :  AT2. 

Cor.  II.  If  two  pyramids  having  equal  altitudes  are 
cut  by  planes  ||  to  the  bases  and  at  equal  distances  from 
the  apices,  the  sections  have  the  same  ratio  as  the  bases. 
Prove  it. 

Cor.  III.  If  in  Cor.  II  the  pyramids  have  also  equal 
bases,  the  sections  are  equal. 


Solid  Geometry. 


169 


Proposition  XXXVII. 

Two  triangular  pyramids   having  equivalent   bases   and 
equal  altitudes  are  equivalent. 

Statement:   If  the   two  pyramids  A  —  BCD   and  A'  — 


Fig.  44- 

B'C'D'  have  equal  bases  (BCD  and  B'C'D')  and  equal 
altitudes  X,  they  are  equivalent. 

Analysis:  By  inscribing  small  equal  prisms  in  each 
pyramid  the  volumes  may  be  compared,  if  the  prisms  be 
made  so  small  that  their  sum  is  practically  equal  to  the 
pyramids. 

Proof:  Divide  the  altitude  into  y  equal  parts  and 
through  the  points  of  division  pass  planes  parallel  to  the 
bases  in  both  pyramids. 

Also,  through  the  intersections  of  these  planes  with  any 
one  face  (in  both  pyramids),  pass  planes  ||  to  the  opposite 
lateral  edge. 

There  will  be  formed  in  both  pyramids  a  series  of 
prisms,  each  prism  in  one  being  equal  to  its  correspond- 


1 70  Solid  Geometry. 

ing  prism  in  the  other;  hence   the   two   series   of  prisms 
are  equal. 

It  is  apparent  that  by  making  the  number  of  parts  into 
which  the  altitude  is  divided,  sufficiently  small,  and  hence 
each  prism  sufficiently  small,  the  aggregate  of  the  prisms 
in  each  pyramid  can  be  made  as  nearly  equivalent  to 
the  volume  of  the  pyramids  as  we  please.  But  the  series 
are  always  equivalent  no  matter  how  small  the  individual 
prisms,  hence  the  pyramids  are  equivalent. 

Proposition  XXXVIII. 

The  volume  oj  a  triangular  -pyramid  is  equal  to  one-third 
the  product  of  its  base  and  altitude. 

Statement:  The  volume  of  the  pyramid  B  —  DEF,  having 
base  DEF  and  altitude,  H  =  J  DEF  X  H. 

Analysis:  If  we  can  express  a  relation  between  pyramids 
and  prisms  we  can  find  the  value  of  this  pyramid  for  we 
know  the  value  of  prisms. 

Prooj:  Through  the  vertices  D  and  F  of  the  base  of  the 
pyramid  draw  CF  and  AD  parallel  and  equal  to  BE. 
Through  A,  B  and  C  pass  a  plane;  also  planes  through 
edges  AD  and  BE,  also  CF  and  BE,  thus  constructing  a 
prism  on  the  base  DEF,  with  an  altitude  equal  to  that  of 
the  pyramid  B  —  DEF. 

This  prism  is  composed  of  the  triangular  pyramid, 
B  —  DEF,  and  the  quadrangular  pyramid  B  —  ADFC. 
Through  A,  B  and  F  pass  a  plane,  forming  triangular 
pyramids  B  -  ADF  and  B  -  AFC. 

Since  ACFD  is  a  parallelogram,  the  diagonal  AF  (the 
line  in  which  the  plane  ABF  intersects  the  face  DFCA) 
divides  this  parallelogram  into  two  equivalent  triangles, 
which  are  the  bases  of  the  two  component  triangular  prisms. 


Solid  Geometry. 


171 


Fig.  45- 


These  triangular  prisms  have 
the  equivalent  triangles  ADF 
and  AFC  for  bases,  and  equal 
altitudes,  because  they  have 
the  same  apex,  and  their 
bases  in  the  same  plane,  hence 
they  are  equivalent. 

Also  the  pyramids  B  -  DEF 
and  F  —  ABC,  have  equal 
bases  (the  bases  of  the  prism) 
and  equal  altitudes,  hence  they 
are  equivalent. 

Now  F  —  ABC  may  be  read 
B  -  ACF  (taking  ACF  as  the 

base  and  B  as  apex,  since  any  face  of  a  triangular  pyra- 
mid may  serve  as  its  base). 

B  -  DEF  =  F  -  ABC  =  B  -  ACF. 
.'.  B  -  ADF  =  B  -  ACF  =  B  -  DEF.     (Axiom.) 

Hence  the  prism  is  divided  into  three  equivalent  pyra- 
mids, of  which  B  —  DEF  is  one. 

Hence,  B  -  DEF  =  J  the  prism  ABC  -  DEF 

and  ABC  -  DEF  =  DEF  X  H  (the  altitude). 

.'.  B  -  DEF  =  J  (DEF  X  H), 

and  since  DEF  is  the  base  of  B  —  DEF,  and  H  is  its  alti- 
tude, the  proposition  is  proved. 


Proposition   XXXIX. 

The  volume  of  any  pyramid  is  equal  to   one-third  the 
product  of  its  base  and  altitude. 
Statement:  Let  A  —  BCDEF  be  a  pyramid  with   base 


172 


Solid  Geometry. 


BCDEF  and  altitude  H,  then 
the  volume  of  ABCDEF= 
BCDEF  X  H. 

Analysis:  By  passing  planes 
through  the  edges  and  diago- 
nals of  the  base  the  pyramid 
is  easily  divided  into  triangular 
pyramids  of  equal  altitude,  that 
are  equal  to  J  the  product  of  the 
base  and  altitude.  Complete 
the  proof  as  in  Prop.  XXXIII. 

Cor.  I.  Pyramids  are  to 
each  other  as  the  product  of 
bases  and  altitudes,  and  if  their 

bases  are  equal  they  are  to  each  other  as  their  altitudes; 

if   their   altitudes    are    equal    they  are   to  each  other   as 

their  bases. 

Cor.  II.     The  volume  of  any  polyhedron  may  be  found 

by  dividing  it  into  triangular  pyramids. 


Proposition  XL. 

The  volume  of  a  jmstrum  of  a  pyramid  is  equal  to  one- 
third  its  altitude  times  the  sum  of  the  areas  of  its  two  bases 
and  their  mean  proportional. 

Statement:  Let  A  -  BCDEF  be  a  frustrum  of  altitude  x 
and  bases  y  and  z;  then  its  volume  =  J  x  (y  +  z  +  \/y.z]. 

Analysis:  A  frustrum  of  a  pyramid  is  evidently  the  dif- 
ference between  two  pyramids,  one  having  the  lower  base 
of  the  frustrum  as  base,  and  the  other,  the  upper  base,  and 
having  the  same  apex  A  of  the  pyramid  from  which 
the  frustrum  was  cut. 

Call  the  volume  of  the  pyramid  A  —  MNOPQ,  v,  and 


Solid  Geometry. 


173 


the    volume     of     the     pyramid 
A  -  BCDEF,  V. 

Let  the  altitude  of  the  pyra- 
mid A  -  MNOPQ  be  T,  then 
volume  of  A  --  BCDEF  = 
J  (x  +  T)  y,  and  volume  of  A— 
MNOPQ  =  }  T  •  2. 

Then  volume  of  the  frustrum 
N  -  BCDEF  =  J  (x  +  T)  y 
-  J  T.3  =   %x.y  + 
J  T    (y  —  z)     (by    rearrange- 
ment). 

y_        (x  +  T)2  1 

z  "  r2 

(Prop.  XXXVI,  Cor.  I.N 


and 
or 

hence, 


Substituting  this  value  of  T  in  the  expression  for  volume 
of  frustrum 


47- 


T 
-TV^ 

T(\/7-Vz")  =  x 

T  =  — — 


7_  (y-z) 


Cor.  Since  ^  x  .  y}  ^  oc  •  z  and  J  ^  Vy  •  2  are  each  the 
volume  of  a  pyramid  with  the  common  altitude  x  and  bases 
y,  z  and  V  y  .  z,  respectively,  how  else  may  proposition  be 
stated  ?  Divide  the  frustrum  into  these  three  pyramids. 


174 


Solid  Geometry. 


Proposition   XLI. 

Two  triangular  pyramids  having  a  trihedral  angle  in 
each  equal,  are  to  each  other  as  the  products  of  the  edges 
including  the  equal  trihedral  angles. 

Statement:  If  A'  -  BCD  and  A  -  EFG  are  two  tri- 
angular prisms  having  the  trihedral  angles  A  and  A'  equal, 
then 

A'  -  BCD  =  A'B  X  A'D  X  A'C 
A  -  EFG        AE  X  AF  X  AG 

Analysis:  As  usual  when  two  geometrical  magnitudes 
are  to  be  compared,  especially  when  they  are  known  to 
have  equal  parts,  superposition  is  the  logical  process. 


Fig.  48. 

Proof:  Combine  the  two  pyramids,  making  A'  coincide 
with  A,  as  to  edges  and  faces,  thus  A'  -  BCD  will  take  the 
position  A  -  BCD  within  A  -  FEG. 

From  B  and  E  drop  perpendiculars  BK  and  EL  to  the 
face  AFG,  and  through  BK,  EL  and  A  pass  a  plane,  inter- 
secting the  face  AFG  in  AKL.  Taking  ACD  as  the  base 
of  A  -  BCD  and  AFG  as  the  base  of  A  -  EFG,  BK  and 


Solid  Geometry.  175 

EL  will  be  their  respective  altitudes,  and  by  Prop.  XXXIX, 
Cor.  I. 

A  -  BCD  =  j  ACD  X  BK  _  ACD  X  BK 
A  -  EFG       JAFG  X  EL  =  AFG  X  EL  ' 
But  ACD^ACXAD 

AFG      AG  X  AF 

X>  Jx          A.JJ 

EL  =  AE' 
hence 

A  -  BCD  =  ACD  X  BK  =  AC  X  AD  X  AB 
A  -  EFG   AFG  X  EL   AG  X  AF  X  AE 


Cor.  Two  tetrahedrons  are  similar  if  three  faces  of  one 
are  similar  respectively  to  three  faces  of  the  other,  and 
are  similarly  placed. 

Definition:  Similar  polyhedrons  are  those  that  have  the 
same  number  of  faces,  respectively  similar  and  placed  in 
the  same  order,  and  their  polyhedral  angles  equal. 

It  follows  easily  from  this  definition  that  similar  poly- 
hedrons may  be  decomposed  into  the  same  number  of 
similar  tetrahedrons;  also  that  the  homologous  edges  of 
similar  polyhedrons  are  proportional,  and  that  any  two 
homologous  lines  have  the  same  ratio  as  two  homo- 
logous edges. 

From  the  theory  of  proportion,  it  follows,  since  the  sur- 
face of  similar  polyhedrons  are  made  up  of  similar  poly- 
gons, that  these  surfaces  are  to  each  other  as  the  squares 
of  any  two  homologous  edges. 


I76 


Solid  Geometry. 


Proposition   XLII. 

The  volumes  of  two  similar  tetrahedrons  are  to  each  other 
as  the  cubes  of  their  edges. 


D 

Fig.  49- 

Statement:  If  A  -  BCD  and  E  —  FGH  are  two  similar 
tetrahedrons,  then 


A-  BCD       AB! 
E  -  FGH 


^3 


or 


AC3  AD; 


or 


:>  etc. 


EG3  EH3' 

Let  the  volume  of  A  -  BCD  be  V,  and  of  E  —  FGH 
beV. 
By  Prop.  XLI, 

X_  =  AB  X  AC  X  AD  =  AB   AC   AD 
V  " "  EF  X  EG  X  EH   EF   EG   EH  ' 

Since  the  faces  are  respectively  similar, 

AB  =  AC  =  AD 
EF      EG      EH ' 

AB       AB       AB  =  Alf  =  AC5  =  AI? 
EF      EF       EF    :  EF3       EG3  ~  EH3 


_V_ 
V 


Solid  Geometry.  177 

Proposition  XLIII. 

The  volumes  of  two  similar  polyhedrons  are  to  each  other 
as  the  cubes  of  any  two  homologous  edges. 

Analysis:  Similar  polyhedrons  can  be  divided  into  similar 
tetrahedrons,  by  planes;  and  since  the  polyhedrons  are  the 
aggregates  of  the  tetrahedrons,  and  the  tetrahedrons  are 
to  each  other,  in  pairs,  as  the  cubes  of  homologous  edges, 
which  are  at  the  same  time  homologous  lines  of  the  similar 
polyhedrons,  the  proof  of  the  proposition  is  evident.  Com- 
plete it. 

EXERCISE    III. 

1.  Any  section  of  a  prism  made  by  a  plane  parallel  to 
an  edge  is  a  parallelogram. 

2.  Show  that  the  diagonals  of  opposite  faces  in  a  par- 
allelopiped  are  parallel. 

3.  The  diagonals  of  a  parallelepiped  bisect  each  other. 

4.  In  a  rectangular  parallelepiped  the  sum  of  the  squares 
of  the  four  diagonals  equals  the  sum  of  the  squares  of  the 
twelve  edges. 

5.  The  volume  of  any  triangular  prism  is  equal  to  half 
the  product  of  any  lateral  face  by  the  perpendicular  to  this 
face  from  the  opposite  edge. 

6.  The  volume  of  any  prism  is  equal  to  the  product  of 
its  lateral  area,  by  half  the  apothem  of  the  base. 

7.  The  square  of  the  diagonal  of  a  cube  equals  three 
times  the  square  of  its  edge. 

8.  If  from  any  point  within  a  regular  tetrahedron  per- 
pendiculars be  dropped  upon  the  four  faces,  the  sum  of 
these  perpendiculars  equals  the  altitude  of  the  tetrahedron. 

9.  Prove  that  a  line  drawn  through  the  point  of  inter- 
section  of  the  diagonals  of  a  parallelopiped  and  termi- 
nating in  opposite  faces,  is  bisected  at  the  point. 


1 78  Solid  Geometry. 

10.  If  the  four  diagonals  of  a  quadrangular  prism  pass 
through  a  common  point  the  prism  is  a  parallelepiped . 

11.  The  base  of  a  prism  is  a  right  triangle,  whose  longest 
and  shortest  sides  are  respectively  17  and  8  in.  ;  the  height 
of  the  prism  is  13  in.     What  is  its  volume? 

12.  The  base  of  a  right  prism  is  a  quadrilateral,  whose 
sides  are  8,  10,  12,  and  14  in.  and  the  lateral  edges  of  the 
prism  are  21  in.     Find  its  lateral  area. 

13.  The    sides    of    the    base    of    a    right     triangular 
prism  are  24,  28,  and  32   and  its  height  is  10.     Find  its 
total  area. 

14.  Two  rectangular  parallelepipeds  have  bases  whose 
dimensions    are    8  X    15    and    9  X   24    respectively.      If 
their  altitudes  are  equal  what  is  the  ratio  of  their  volumes  ? 

15.  The  dimensions  of  two  rectangular  parallelepipeds 
are    respectively  5,   9,  16,    and   9,    12;   the   ratio   between 
their  volumes  is  i:  3.     What  is  the  third  dimension  of  the 
second  ? 

16.  Find  the  dimensions  of  a  cube  equal  in  volume  to 
a  parallelepiped  whose  dimensions  are  27,  28,  and  35. 

17.  If  the  dimensions  of  a  brick  are  8  X  4  X  3  in.  how 
many  bricks  will  it  require  for  a  wall  8  in.  thick,  30  ft.  long 
and  20  ft.  high  ? 

18.  A  stack  has  the  form  of  a  frustrum  of  a  cone.     The 
diameter  of  its  lower  base  is  6  ft.  and  of  its  upper  base  4  ft. 
If  it  is  50  ft.  high,  what  will  it  cost  to  paint  it  at  7  cents  per 
square  yard? 

19.  What  will  it  cost  to  excavate  a  trench  4  ft.  deep, 
6  ft.  wide'  at  the  top,  4  ft.  wide  at  the  bottom,  and  50  ft. 
long,  at  1 6  cents  per  cubic  yard? 

20.  Taking  231  cu.  in.  to  the  gallon,  how  much  water 
will  a  bucket  hold,  if  it  is  12  in.  high,  has  a  bottom  diam- 
eter of  8  in.,  and  a  top  diameter  of  12  J  in.  ? 


Solid  Geometry.  179 

21.  How  much  slate  will  it  take  to  cover  a  tower  in  the 
form  of  a  regular  quadrangular  pyramid,  24  ft.  high  and 
with  a  base  10  ft.  on  a  side,  if  the  slate  averages  12  X  8  in. 
and  6  in.  is  uncovered? 

22.  A  section  of  a  tunnel  is  a  rectangle  8  X  12  ft.,  sur- 
mounted by  a  semi-circle  whose  diameter  is  the  smaller 
dimension  of  the  rectangle.     If  the  tunnel  is  f  of  a  mile 
long,  how  many  cubic  yards  of  dirt  were  removed  ? 

23.  A  log  is  15  ft.  long,   16  in.  diameter  at  the  larger 
end  and   12   in.   diameter  at  the  smaller.      What  will  be 
the  dimensions  of  the  largest  square  section  pillar  that  can 
be  sawed  from  it,  and  how  many  cubic  feet  of  waste  will 
there  be? 

24.  The  areas  of  the  bases  of  a  frustrum  of  a  pyramid 
are  48  and  27  sq.  ft.,  respectively,  and  its  altitude  is  9   ft. 
What  is  the  altitude  of  the  completed  pyramid? 

25.  The  altitude  and  lateral  edge  of  a  frustrum  of  a 
regular  triangular  pyramid  are  4  and  5,  respectively,  and 
each  side  of  its  upper  base  is  \/3 .     Find  its  volume. 

26.  The  bases  of  frustrum  of  a  pyramid  are  rectangles 
whose  sides  are  27  and  15,  and  9  and  5,  respectively,  and 
the  line  joining  their  centres  is  perpendicular  to-  both  and 
equal  to  12.     Find  lateral  area  and  volume. 

27.  The  lateral  surface  of  a  pyramid  is  greater  than  its 
base. 

28.  The   altitude   of   a   pyramid  is   divided   into   three 
equal  parts  by  planes  parallel  to  its  base?     Find  the  ratio 
between  the  three  solids  thus  formed. 

29.  The  volume  of  a  cube  is  iff  cu.  ft.     Find  the  length 
of  its  diagonals. 

30.  The  lines  joining  each  vertex  of  a  tetrahedron  with 
the  point  of  intersection  of  the  medial  lines  of  the  opposite 
faces  all  meet  in  a  common  point,  which  divides  each  line 


180  Solid  Geometry. 

in  the  ratio  i  :  4  (this  point  being  the  centre  of  gravity  of 
the  tetrahedron).     Prove  it. 

31.  What  is  the  edge  of  a  cube  whose  entire  surface  is 
63-375  square  feet? 

32.  On  three  given  lines  intersecting  in  space,  construct 
a  parallelepiped. 

33.  The  altitude  of  a  pyramid  is  64  in.     Divide  the 
pyramid  into  three  equal  parts  by  planes  parallel  to  the  base, 
by  determining  the  points  on  the  altitude,  where  the  planes 
cut  them. 

34.  A  section  of  a  tetrahedron  made  by  a  plane  parallel 
to  two  non-intersecting  edges  is  a  parallelogram.     Prove  it. 

35.  The  altitude  of  a  pyramid  is  8  ft.  and  its  base  is  a 
regular  hexagon  with  4  foot  sides.     Find  volume. 

36.  The  slant  height  of  the  frustrum  of  a  regular  pyramid 
is  15  in.;  the  sides  of  its  square  bases  are  respectively  30 
and  12  in.     Find  volume  of  frustrum. 

37.  Within  a  4o-foot  steeple,  in  the  form  of  a  quadran- 
gular pyramid  with  square  base,  a  platform  is  to  be  built 
25  ft.  from  its  base.     If  the  base  of  the  steeple  is  16  ft.  on 
a  side,  how  many  square  feet  of  lumber  will  be  required 
for  the  platform? 

38.  Around  the  same  steeple  at  a  height   of  28  ft.   a 
platform  3  ft.  wide  is  to  be  built.     How  many  square  feet 
of  lumber  will  be  required  ? 

39.  A  wagon  bed  is  5  X  17  ft.  at  the  top,  and  4  X  16  ft. 
at  bottom,  corresponding  dimensions.     If  it  is  4  ft.  deep, 
how  many  bushel  of  coal,  counting  2,688  cubic  inches  to 
the  bushel,  will  it  hold? 

40.  A  square  chimney  is  120  ft.  9  in.  high;  its  width  af: 
the  base  is  10  ft.  9  in.,  and  at  the  top,  5  ft.  9  in.     The 
cavity  is  a  square  prism  2  ft.  6  in.  on  the  sids.     How  many 
cubic  feet  of  masonry  in  the  chimney? 


Solid  Geometry.  181 

CYLINDERS. 

Def.  I.  A  cylindrical  surface  is  one  generated  by  a  line 
moving  parallel  to  a  fixed  straight  line  and  always  touch- 
ing a  fixed  curve,  not  in  the  same  plane  as  the  fixed  straight 
line. 

The  moving  line  is  called  the  generatrix,  and  the  fixed 
curve  is  called  the  directrix. 

The  generatrix  in  any  position  is  called  an  element  of  the 
cylindrical  surface. 

Def.  II.  A  cylinder  is  a  solid  bounded  by  a  cylindrical 
surface  with  a  closed  directrix  and  two  parallel  sections  of 
it.  Compare  with  a  prism. 

The  parallel  sections  are  called  the  bases,  and  the  cylin- 
drical surface,  the  lateral  surface. 

Def.  III.  The  altitude  of  a  cylinder  is  the  perpendicular 
distance  between  its  bases. 

Def.  IV.  A  right  cylinder  is  one  whose  elements 
are  all  perpendicular  to  the  planes  of  the  bases.  In 
other  words  it  is  one  whose  elements  are  all  equal  to  the 
altitude. 

Def.  V.  A  circular  cylinder  is  one  whose  bases  are 
circles.  What  must  be  the  form  of  the  directrix? 

Def.  VI.  It  is  evident  that  a  right  circular  cylinder 
may  be  generated  by  a  rectangle,  revolving  about  one  of 
its  sides  as  an  axis.  Such  a  cylinder  is  often  called  a 
cylinder  of  revolution. 

By  the  word  cylinder  is  usually  meant  a  circular  cylinder. 

Def.  VII.  The  line  joining  the  centres  of  the  bases  is 
called  the  axis  of  a  cylinder. 

Def.  VIII.  A  plane  is  tangent  to  a  cylinder,  when  it 
contains  only  one  element. 

Def.  IX.  A  prism  is  said  to  be  inscribed  in  a  cylinder, 
when  its  bases  are  inscribed  in  the  bases  of  the  cylinder, 


182 


Solid  Geometry. 


and  its  lateral  edges  are  elements  of  the  cylinder.     Define 
circumscribed  prisms. 

Remarks:  It  is  perfectly  clear  that  the  bases  of  a  cylinder 
are  equal;  that  any  section  parallel  to  the  bases  is  equal 
to  them;  and  that  sections  that  are  parallel  and  cut  all  the 
elements  are  equal. 

Proposition   XLIV. 

Any  section  of  a  cylinder  made  by  a  plane 
passing  through  an  element  is  a  parallelogram. 

Statement:  If  abed  is  such  a  section  of  the 
cylinder  ad,  it  is  a  O. 

Analysis:  The  definition  of  a  O  requires  a 
proof  that  the  sides  of  abdc  are  ||. 

Proof:    Let    bd    be    the    element    through 
which    the    section   is    drawn.      Through    a 
draw   an    element  of   the  cylinder;   this   ele- 
ment lies   in    the   plane   abdc    (Why?),    and, 
being  an  element,  also  lies  in  the  surface  of 
Fig.  so.        ^e    Cyimc}erj    hence    is    the   intersection    of 
abdc  with  the  surface,  hence  it  coincides  with   ac.     But   it 
is  parallel  to  bd,  since  they  are  both  elements. 

/.  ac  is  ||  to  bd.     Also  bd  is  ||  to  cd.  (Why?) 

.*.  abdc  is  a  O. 

Proposition   XLV. 

The  lateral  area  of  a  circular  cylinder  is  equal  to  the 
product  of  the  circumference  of  a  right  section  by  an  element. 

Statement:  Let  cd  be  a  right  section  and  ab  an  element 
of  the  cylinder  ae,  then  the  lateral  area  S  =  circumference 
of  cd  X  ab,  say,  S  =  C  X  ab. 

Analysis :  The  proposition  immediately  suggests  the 
corresponding  proposition  on  the  prism,  and  as  it  is  readily 


Solid  Geometry. 


183 


Fig.  si. 


shown  that  a  cylinder  is  nothing  but  a  prism  with  an  infi- 
nite number  of  faces,  we    are   led   to   in- 
scribe    a    prism    and    increase    its    faces 
indefinitely. 

Proof:  Inscribe  a  regular  prism  of  any 
number  of  faces  in  the  cylinder;  call  its 
lateral  area,  S',  its  lateral  edge,  ab,  and  the 
perimeter  of  its  right  section  inscribed  in 
the  right  section  of  the  cylinder,  P.  Then 
S'=  P  X  ab. 

But    if    the    number    of    faces    of    this 
prism   be   indefinitely  increased,    say    by 
successive  doubling  of   the   number  of  sides,  it    will  ulti- 
mately coincide  with  the  cylinder.     Then 

S'  =  S  and  P  =  C. 
.';  S  =  C  X  ab. 

Remark:  Similarly  it  may  be  proved  that  the  volume  of 
a  cylinder  equals  the  product  of  its  base  and  altitude.  In 
fact  all  the  propositions  applying  to  prisms  apply  equally 
to  cylinders. 

Proposition   XLVI. 

The  lateral  or  entire  areas  of  similar  cylinders  of  revolu- 
tion are  to  each  other  as  the  squares  of  their  altitudes  (or  an 
elementy  or  as  the  squares  of  their  radii,  and  their  volumes 
are  to  each  other  as  the  cubes  of  their  altitudes  or  radii. 

Statement:  Let  A  and  B  be  two  cylinders  of  revolution; 
S  and  S'  their  lateral  surfaces;  T  and  T'  their  total  surfaces; 
V  and  V  their  volumes;  R  and  R'  their  radii;  and  H  and 
H'  their  altitudes,  then  S  :  S'  :  :  R2  :  R'3  :  :  H2  :  H/2; 
T  :  T'  :  :  R2  :  R'2  :  :  H2  :  H/2  and 

V:  V':  :R3:R/3:  :  H3  :  H'3. 


184 


Solid  Geometry. 


Analysis:  Being  cylinders  of  revolution  they  are  gen- 
erated by  rectangles,  which,  of  course,  are  similar,  and 
since  the  radii  and  altitudes  constitute  the  two  dimensions 

of  these  rectangles  their  ra- 
tios are  equal. 

Proof:  Since  the  lateral 
areas  of  cylinders,  like 
prisms,  equal  the  perimeter 
of  a  right  section  multi- 
plied by  an  element,  and 
these  being  right  cylinders, 
the  bases  are  right  sections 


and  the  altitudes  equal  the 
elements, 


Fig.  52. 


S   =  2  TrRH  and  S'  =  2  TrR'H',  T  =  2  TrR2  +  2  TrRH, 
T'  =  2  TrR'2  +  2  TrR'H',  V  -  TrR2H  and  V  -  TrR'2H'. 

[From  like  propositions  on  prisms.] 
(The  total  area  includes  the  two  bases  with  the  lateral 
surface  and  the  volume  equals  the  product  of  base  by 
altitude.) 

S   ._   2  TrRH          RH        R        H 
S'       2  TrR'H'        R'H'       R'       H'' 

and  since  the  rectangles  are  similar 

*.  .  H.    hence  5.- JL,  x  *, -£  x  *- -  *  , 

R        t±  o         R        JLL        R         R        R 


or 


H2 


Hf      H' 

T^  =  2  TrR2  +  2  TrRH  R2+RH         R  (R  +  H) 

Tf       2  TrR'2+  2  TrR'H'       R'2+  R'H'      R'  (R'+  H') 


R 
R' 


R'  +  H 


Solid  Geometry.  185 

By  theory  of  proportion  if 
R  :  R'  :  :  H  :  H',  then  R  +  H  :  R'  +  H'  :  :  R:  R':  :  H  :  H'. 

I  =  JL  y  A-zrBi-JL  x  —   =  51 
'   T       R'       R'       R/2       H'       H'      H/2  ' 

Again : 

_V_        ?rR2H    =  R2         H   =  R3   =  H3 

V'       7iR'2H'      R'2       H'      R'3      H/3  ' 
CONES. 

Def.  I.  A  conical  surface  is  one  generated  by  a  line 
which  moves,  always  touching  a  fixed  curved  line  and 
passes  through  a  fixed  point,  not  in  the  plane  of  the  curve. 

The  terms,  generatrix  and  directrix  have  the  same  sig- 
nificance as  in  the  cylindrical  surface,  and  the  fixed  point 
is  called  the  vertex  or  apex. 

Def.  II.  If  the  generatrix  does  not  terminate  in  the 
fixed  point,  it  generates  two  portions,  called  the  upper 
and  lower  nappes,  respectively. 

Def.  III.  The  generatrix  in  any  position  is  called  an 
element  of  the  conical  surface. 

Def.  IV.  A  cone  is  a  solid  bounded  by  a  conical  surface 
with  a  closed  curve  for  directrix,  and  a  section  cutting  all 
the  elements. 

The  base,  total  and  lateral  surface  have  the  usual  meaning. 

Def.  V.  The  altitude  of  a  cone  is  the  perpendicular 
distance  from  the  apex  to  the  plane  of  the  base. 

Def.  VI.  A  circular  cone  is  one  having  a  circular  base. 
The  straight  line  joining  the  apex  and  the  centre  of  the 
base  is  called  the  axis. 

Def.  VII.  If  the  axis  is  perpendicular  to  the  base,  it 
is  a  right  cone;  otherwise,  it  is  called  an  oblique  cone. 

Def.  VIII.  A  cone  of  revolution  is  a  right  circular  cone 
generated  by  a  right  triangle  revolving  about  one  of  its 


i86 


Solid  Geometry. 


legs.    The  hypotenuse  of  this  triangle  in   any  position  is 
called  the  slant  height. 

Similar  cones  of  revolution  are  those  generated  by  similar 
right  triangles. 

Def.  IX.  A  tangent  line  to  a  cone  is  a  straight  line, 
which  touches  the  surface  in  a  single  point. 

Def.  X.  A  tangent  plane  is  one  containing  one  element 
of  the  surface,  known  as  the  element  oj  contact. 

Def.  XL  A  pyramid  is  said  to  be  inscribed  in  a  cone, 
when  its  base  is  inscribed  in  the  base  of  the  cone.  A 
pyramid  is  circumscribed  about  a  cone  when  its  base  is 
circumscribed  about  the  base  of  the  cone  and  its  faces  are 
tangent  to  the  conical  surface. 

Def.  XII.  A  truncated  cone  is  the  portion  of  a 
cone  comprised  between  the  base  and  any  plane  sec- 
tion. If  the  section  is  parallel  to  the  base  it  is  called  a 
jrustrum. 

The  altitude  of  a  frustrum  is  the  perpendicular  distance 
between  its  bases,  and  the  slant  height  of  a  frustrum  of  a 
cone  of  revolution  is  the  altitude  of  one  of  its  trapezoid 
faces. 

Proposition   XL VII. 

Every  section  of  a  cone  made  by  a  plane 
through  the  vertex  is  a  triangle. 

Statement:  Let  ABC  be  a  section  of  the 
cone  A  —  BCD  made  by  the  plane  MN 
through  A,  then  ABC  is  a  triangle. 

Analysis:  The  lines  of   intersection   of 
the  plane  MN  with  the  conical  surface 
determines    the    character  of  the  figure. 
The  intersection  with  the  base,  which  is 
Fig.  53.  a  plane,  we  know  to  be  a  straight  line. 

Since    the    elements    are    straight    lines,  meeting   at   A, 


Solid  Geometry,  187 

it  is  only  necessary  to  show  that  the  other  two  sides  are 
elements. 

Proof:  Let  BC  be  the  intersection  of  the  cutting  plane 
with  the  base,  then  it  is  a  straight  line.  Draw  the  straight 
lines  AB  and  AC. 

They  are  elements.  (Why?) 

They  also  lie  in  the  cutting  plane.  (Why?) 

.'.  they  are  intersections  of  the  plane  with  the  conical 
surface. 

/.  the  section  ABC  is  a  triangle. 

Proposition  XLVTII. 

Every  section  of  a  cone  parallel  to  the  base  is  a  circle. 

Statement:  Let  FKH  be  a  section  of  cone  A  —  EDG  || 
to  base  EDG.  Then  FKH  is  a  circle. 

Analysis:  If  FKH  is  a  circle,  the  points 
on  its  circumference  are  all  equally  distant 
from  some  point  within  it.  If  it  is  a  circle  it 
is  similar  to  the  base,  and  the  axis  of  the  cone 
ought  to  pass  through  this  central  point. 

By  connecting,  then,  the  point  in  FKH, 
where  the  axis  cuts  it  with  any  two  or  three 
points  on  its  circumference  and  comparing, 
by  the  use  of  lines  in  the  base,  the  truth  of 
the  proposition  may  be  seen.  Fis-  54- 

Proof:  Draw  the  axis,  AC,  cutting  Fkn  in  B.  Draw 
also  the  elements  AE,  AD  and  AG,  touching  FKH  at  F, 
K  and  H,  respectively.  Join  B  with  F  and  K.  Then  the 
triangles  ABF  and  ACE  are  similar,  also  ABK  and  ACD. 

(Why?) 


1 88  Solid  Geometry. 

:.  FB  :  EC  :  :  AB  :  AC  and  BK  :  CD  :  :  AB  :  AC, 
.'.  FB  :  EC  :  :  BK  :  CD  but  EC  =  CD 

(radii  of  circle  EDG.) 
.'.  FB  =  BK. 

Hence  FKH  is  a  circle,  with  centre  at  B. 

The  following  propositions  may  be  immediately  inferred 
from  the  fact  that  a  cone  is  merely  a  pyramid  of  an  infinite 
number  of  faces :  — 

First.  The  lateral  area  of  a  cone  of  revolution  (corre- 
sponding to  a  regular  pyramid}  is  equal  to  one-half  the 
product  of  the  slant  height  by  the  circumference  of  the 
base. 

Hence  if  S  represents  lateral  area,  T  total  area,  L  the 
slant  height,  and  R  the  radius  of  the  base,  in  a  cone  of 
revolution,  then 

S  =  i  (2  TrR  X  L)  =  TrR  X  L. 

T  =          TrR  X  L    +  TrR2  =  TrR  (L  +  R). 

Second.  The  volume  of  a  circular  cone  equals  one-third 
the  product  of  the  altitude  by  the  area  of  the  base;  hence, 
calling  V  the  volume,  H  the  altitude,  and  R  the  radius  of 
the  base, 

V  =  J  TrR2H. 

Third.  The  lateral  area  of  the  frustrum  of  a  cone  of 
revolution  equals  one-half  the  sum  of  the  circumferences 
of  upper  and  lower  base,  multiplied  by  the  slant  height; 
or  what  amounts  to  the  same,  the  lateral  area  of  such  a  frus- 
trum equals  the  slant  height  by  the  circumference  of  the 
mid-section. 

Fourth.  The  volume  of  the  frustrum  of  a  circular  cone 
is  equivalent  to  one-third  of  the  altitude  multiplied  .by  the 
sum  of  the  upper  base,  the  lower  base,  and  a  mean  propor- 
tional between  them. 


Solid  Geometry.  189 

Proposition  XLIX. 

The  lateral  areas,  or  the  total  areas  of  two  similar  cones 
of  revolution  are  to  each  other  as  the  squares  of  their  alti- 
tudes, as  the  squares  of  their 
radii,  or  as  the  squares  of  their 
slant  heights;  and  their  vol- 
umes are  as  the  cubes  of  these  L/ 
like  dimensions. 

Call  S  and  S',  the  lateral 
surfaces;  T  and  T',  the  total 
surfaces;  V  and  V,  the  vol- 
umes; H  and  H',  the  alti-  *'*•  55- 

tudes;  R  and  R/,  the  radii;    L   and  L',  the  slant  heights; 
then 

S    ==  TrRL,      S'  =  TrR'L',     T  =  TrR  (R  +  L). 
T'  =  TrR'  (R'  +  I/),     V  =  J  ;rR2H 

and  V  =  J  7rR'2H' 

S          RL          RL        R       L 


-        ?rR(R  +  L)        Jl        R  +  L 
T'      7rR'(R'  +  I/)    "   R'      R'+L'' 

.=  R*        H_ 
' 


V7      j7rR/2H'      R/2       H' 

T)  TT  T  T)        I       T 

But  -*L  =  *L  =  A  =    K  +  L.    . 

R'       H'       L'       R'  +  L' 

(Since  the  generating  triangles  are  smiliar.) 

"    §"'  =  ~R'  X  ~Rf  =  R7"2  =  H7"2  =  L72 
T       R2  V       R3 


190  'Solid  Geometry. 

EXERCISE    IV. 

1.  The  total  area  of  a  cylinder  of  revolution  is  385  sq. 
in.    The  height  is  four  times  the  radius  of  the  base.     Find 
radius  and  height. 

2.  How  much  sheeting  will  it  take  for  a  smoke  stack 
60  ft.  high  and  12  in.  in  diameter? 

3.  The  altitude  of  a  cone  of  revolution  is  15  in.,  the 
radius  of  its  base  is  8  in.     What  is  its  lateral  surface? 

4.  The  sloping  face   of   a  conical   excavation  measures 
13  ft.,  and  it  is  10  ft.  across  at  the  mouth.     What  is  the 
amount  of  excavation  ? 

5.  A  torpedo  is  made  up  of  a  right  cylinder,  capped  by 
two  right  cones.     The  cylindrical  part  is  4  ft.  long  and  the 
conical  parts  each  12  in.  high.     The  diameter  is  zoj  in. 
Find  contents. 

6.  The  bell-shaped  mouth  of  a  cylindrical  stack,  14  in. 
in  diameter,  is  18  in.  long,  measured  on  its  sloping  face, 
and  its  largest  diameter  is  20  in.     How  much  sheeting  will 
it  take  ? 

7.  Find  the  volume  of  a  cone  of  revolution  whose  slant 
height  is  26  ft.  and  whose  lateral  area  is  943  sq.  ft. 

8.  A  cubical  mass  of    lead  whose  total  surface  is  486 
sq.  in.  is  melted  and  cast  into  a  cone,  the  diameter  of  whose 
base  is  7  in.     What  will  be  the  height  of  the  cone  ? 

9.  A  tapering  discharge  pipe  f  in.  thick,  whose  largest 
diameter  is  4  in.  and  smallest  2  in.,  is  10  ft.  long.     How 
much  metal  was  required  for  it  ? 

10.  How  far,  respectively,   from  the  vertex  of  a  cone 
are  two  sections  parallel  to  the  base,  if  they  divide  the  cone 
into  three  equal  parts,  the  altitude  of  the  cone  being  9  ft. 

11.  A  right  cone  is  cut  into  two  unequal  parts  by  a 
plane  parallel  to  the  base.     The  lower  part  is  7  times  the 


Solid  Geometry.  191 

upper  part,  and  the  height  of  the  cone  is  12  in.     Find 
distance  from  the  vertex  to  the  plane. 

12.  A  bucket  has  an  upper  diameter  of  15  in.,  a  bottom 
diameter  of  12  in.,  and  a  height  of  14  in.     Counting  231  cu. 
in.  to  a  gallon,  find  the  contents  in  gallons. 

13.  The  lateral  area  of  a  cone  is   1,232  sq.  ft.     The 
slant  height  is  25  ft.      What  is  the  area  of  a  section  6  ft. 
from  the  vertex  ? 

14.  The  radius  of  the  base  of  a  cone  is  }  the  height, 
and  the  lateral  surface  is  330  sq.  in.     Find  the  altitude 
and  the  radius. 

15.  The   contents   of   a   cylindrical    cistern   in    gallons 
(231  cu.  in.  to  the  gallon)  is  given  by  the  formula,  D2  X  H 
X  -0034,  where  D  is  the  diameter  of  the  base,  and  H  is 
the  altitude.     Verify  the  formula. 

16.  The  total  areas  of  two  similar  cylinders  of  revo- 
lution are  50  and  72  sq.  ft.,   respectively.     The  volume 
of    the    larger   is    1,728  cu.  ft.      What  is  the  volume  of 
the  other? 

THE    SPHERE. 
Definitions. 

Def.  I.  A  sphere  is  a  solid  bounded  by  a  surface,  every 
point  of  which  is  equally  distant  from  a  point  within, 
called  its  centre. 

Evidently  a  sphere  may  be  generated  by  the  revolution 
of  a  semi-circle  about  its  diameter. 

Def.  II.  The  radius  of  a  sphere  is  the  distance  from  its 
centre  to  any  point  on  its  surface.  A  diameter  is  a  line 
through  the  centre  limited  by  the  surface.  It  is  equal  to 
twice  the  radius. 

Def.  III.  A  line  or  plane  is  tangent  to  a  sphere,  when 
it  has  but  one  point  in  common  with  the  sphere. 


192  Solid  Geometry. 

Def.  IV.  A  great  circle  of  a  sphere  is  a  circle  whose 
plane  divides  the  sphere  into  two  equal  parts.  All  other 
circles  are  known  as  small  circles.  The  plane  of  a  great 
circle  passes  through  the  centre. 

Def.  V.  The  axis  of  a  circle  of  a  sphere  is  that  diameter 
of  the  sphere  that  is  perpendicular  to  the  plane  of  the 
circle.  Its  ends  are  called  poles  of  the  circle. 

Def.  VI.  The  distance  between  two  points  on  a  sphere 
is  understood  to  be  the  arc  of  a  great  circle  included  be- 
tween them. 

This  distance  is  the  shortest  distance  between  the  two 
points;  hence  the  designation. 

Def.  VII.  The  distance  on  the  spherical  surface  from 
the  nearer  pole  of  a  small  circle  to  any  point  on  its  cir- 
cumference is  called  the  circle's  polar  distance. 

Solids  are  said  to  be  inscribed  in  a  sphere,  when  all  their 
vertices  are  in  the  spherical  surface. 

Proposition  L. 

All  sections  of  a  sphere  are  circles. 

Analysis:  The  characteristic  of  a  circle  is  that  its  points 
shall  all  be  equally  distant  from  a  point  within.  Since  we 
know  that  all  radii  of  the  sphere  are  equal,  it  is  suggested 
to  connect  these  points  with  the  centre  of  the  sphere,  and 
to  locate  the  centre  of  the  section  by  drawing  a  diameter 
perpendicular  to  it,  since  it  can  be  inferred  from  what  we 
know  of  circles  that  a  diameter  perpendicular  to  this  sec- 
tion, will  bisect  all  the  lines  through  its  foot,  in  the  section. 
Complete  the  proof. 

As  corollaries,  the  following  may  be  inferred :  — 

(a)  The  line  joining  the  centre  of  a  sphere  to  the  centre 
of  any  of  its  circles,  is  perpendicular  to  the  plane  of  the 
circle. 


Solid  Geometry.  193 

(b)  Circles  of  a  sphere  equally  distant  from  its  centre 
are  equal. 

(c)  Of   two  circles  made   by  planes  unequally  distant 
from  the  centre,  the  nearer  is  the  greater. 

(d)  All  great  circles  of  a  sphere  are  equal. 

(e)  Two  great  circles  bisect  each  other. 

(/)  If  the  planes  of  two  great  circles  are  perpendicular 
each  passes  through  the  poles  of  the  other. 

(g)  Through  any  three  points  on  the  surface  of  a  sphere 
one  circle  may  be  passed.  One-fourth  of  a  great  circle  is 
called  a  quadrant. 

Proposition  LI. 

A  point  on  the  surface  of  a  sphere  which  is  at  a  quad- 
rant's distance  from  each  of  two  other  points,  not  the  extremi- 
ties of  a  diameter,  is  a  pole  of  the  great  circle  passing  through 
these  points. 

Statement:  Let  A  on  the  sphere  D,  be  at  a 
quadrant's  distance  from  B  and  C,  then  A  is  the 
pole  of  the  great  circle  BCE  through  B  and  C. 

Analysis:  Since,  by  definition,  the  pole  of 
a  circle  is  the  end  of  the  diameter  perpendic- 
ular to  its  plane,  it  is  immediately  suggested 
to  draw  through  A  a  diameter  and  prove  it 
perpendicular  to  the  plane  of  EEC,  remembering  that  a 
line  is  perpendicular  to  a  plane  if  it  is  perpendicular  to 
two  lines  in  that  plane,  drawn  through  its  foot. 

Proof:  Draw  the  diameter  AF,  cutting  the  plane  EBC 
at  D  (since  the  plane  of  a  great  circle  must  pass  through 
the  centre  of  the  sphere). 

Draw  the  radii  DB  and  DC.  Then  it  remains  to  show 
that  A  ADB  and  ADC  are  right  angles. 

Complete  the  proof,  remembering  that  arcs  AC  and  AB 
are  quadrants. 


i94 


Solid  Geometry. 


Proposition  LII. 

Given  a  material  sphere,  to  -find  its  diameter. 
Analysis:  Since  it  is  impossible  to  enter  the  sphere,  it 
is  necessary  to  take  points  on  its  surface,  and  from  super- 
ficial measurements  to 

H 

construct  sections,  and 
by  using  our  knowl- 
edge of  the  relations 
of  lines  within  a 
sphere,  find  our  di- 
ameter. 

Construction:  Let  O 
be  the  sphere.  With 
any  point,  A  on  the 
sphere  as  a  centre, 
describe  a  circle  on  the 
surface;  take  any  three 
points  D,  C  and  E  on 
it,  and  with  the  dividers  measure  the  distances  (chords) 
DC,  DE  and  EC. 

Draw  the  triangle  DCE  formed  of  these  chords. 
Find  the  centre  B  of  this  triangle;  B  will  be  the  centre 
of  the  small  circle  passing  through  E,  C  and  D,  and  BC  is 
its  radius. 

Construct  a  right  triangle  with  AC  (measured  with  the 
dividers)  as  hypotenuse  and  BC  as  one  leg;  let  ABC  be 
this  triangle.  Draw  a  perpendicular  to  AC  at  C,  and 
produce  AB  to  meet  it  at  F,  then  AF  is  the  required  diameter. 
For  A  is  the  pole  of  the  circle,  ECD,  and  BC  its  radius, 
and  since  ABF  was  drawn  perpendicular  to  BC  at  B,  the 
centre  of  the  circle  ECD,  it  passes  through  the  centre  of 
the  sphere.  (Why?) 


Fig.  57- 


Solid  Geometry.  195 

Also  ACF  is  a  right  angle  and  hence  inscribed  in  a  semi- 
circle ;  hence  its  sides  AC  and  CF  terminate  at  the  ends  of 
the  diameter  at  A  and  F. 

The  same  thing  may  be  shown  by  comparison  with  a 
section  of  the  sphere  through  its  diameter. 

The  triangle  ACF  is  then  seen  to  be  identical  with  its 
prototype  ACF  in  the  sphere. 

Proposition  LIU. 

A  plane  perpendicular  to  a  radius  at  its  extremity  is 
tangent  to  the  sphere  at  that  point. 


Fig.  58. 


Statement:  If  the  plane  xy  is  perpendicular  to  a  radius 
OB  of  the  sphere  O,  at  B,  it  is  tangent  to  the  sphere  at  B. 

Analysis:  Since  a  plane  is  tangent  when  it  has  but  one 
point  in  common  with  the  sphere,  if  a  line,  other  than  the 
radius  OB,  be  drawn  from  O  to  any  point  of  xy  other  than 
B,  it  can  be  shown  that  it  is  longer  than  OB  and  hence 
outside  the  sphere.  Complete  the  proof. 

Cor.  I.  A  plane  tangent  to  a  sphere  is  perpendicular 
to  the  radius  drawn  to  the  point  of  contact. 

Cor.  II.     A  line  tangent  to  a  sphere  at  a  given  point, 


196 


Solid  Geometry. 


lies  in  the  plane  tangent  to  the  sphere  at  that  point,  and 
conversely. 

Cor.  III.     The  plane  of  two  lines  tangent  to  the  sphere 
at  the  same  point  is  tangent  to  the  sphere  at  that  point. 

Proposition  LIV.      (Problem.) 

1  o  inscribe  a  sphere  in  any  given  tetrahedron. 
Analysis:  That  a  sphere  may  be  inscribed  in  the  tetra- 
hedron A  —  BCD,  its  centre  must  be  so  determined  that 
it  shall  be  equally  distant  from  the  four  faces.     In  that 
connection,  the  locus  of  points  equally  distant  from  the 

faces  of  a  dihedral  angle 
is  suggested. 

Construction:  Bisect  the 
dihedral  angles  BD,   CA 
and  AB  by  planes,  inter- 
secting at    O.     Then    O 
will  be  the  required  centre. 
Proof:  The  plane  BOD 
D   is  the  locus  of  all  points 
equally  distant  from  the 

faces  ABD  and  BCD,  the  plane  OCA  is  the  locus  of  all 
points  equally  distant  from  the  faces  ACD  and  ACB,  and  the 
plane  OAB  is  the  locus  of  all  points  equally  distant  from 
the  faces  ABD  and  ACB.  Therefore,  the  points  O,  where 
these  planes  intersect,  is  equally  distant  from  all  the  faces. 
The  planes  must  intersect  in  a  common  point  as  O, 
because  O  is  the  only  point  that  possesses  the  characteristic 
property  of  all  the  bisecting  planes. 


Solid  Geometry. 


197 


Proposition  LV.     (Problem.) 
To  circumscribe  a  sphere  about  any  tetrahedron. 
Analysis:  Since  the  centre  of  such  a  sphere  must  be 
equally  distant  from  the  four  vertices  of  the  tetrahedron, 
it  is  suggested  that  a  perpendicular  to  a  plane  is  the  locus 
of  all  points  equally  distant  from  points  in  the  plane,  equally 
distant  from  its  foot.     Hence  a  perpendicular  to  the  plane 
of  any  of  the  triangular  faces   at  its  centre  fulfills  the 
condition. 

Complete  the  proof. 

Proposition  LVI. 

The  intersection  of  two  spherical  surfaces  is  the  circum- 
ference of  a  circle  whose  plane  is  perpendicular  to  the  line 
joining  the  centres  of  the  surfaces  and  whose  centre  is  in 
that  line. 

Statement:  If  the  surfaces  of  the  spheres  A  and  B  inter- 
sect, to  prove  that  the 
line  of  intersection  is  the 
circumference  of  a  circle , 
whose  centre  lies  in  the 
line  AB. 

Analysis:  Since  a 
sphere  may  be  generated 
by  revolving  a  semi-circle 
about  its  diameter,  the  Fig-  59a- 

relations  between  the  lines  can  be  determined  by  making 
a  section  through  the  two  centres. 

Proof:  Pass  a  section  through  the  two  centres,  intersect- 
ing the  two  surfaces  in  the  great  circles  A  and  B,  which 
intersect  at  C  and  D.  Join  C  and  D  by  a  straight  line 
also  A  and  B. 


198  •  Solid  Geometry. 

By  Plane  Geometry  AB  is  perpendicular  to  CD  at  the 
middle  point  F.  Hence  if  this  whole  plane  figure  be  re- 
volved about  AB  as  an  axis,  the  points  C  (or  D)  will 
describe  a  circle  with  centre  at  F  on  the  line  AB. 

Definitions. 

The  angle  formed  by  two  intersecting  curves  is  the  angle 
formed  by  the  tangents  to  these  curves  at  the  point  of 
intersection. 

If  the  curves  are  arcs  of  great  circles,  the  angle  is  called 
a  spherical  angle. 

A  spherical  polygon  is  a  portion  of  the  surface  of  a  sphere, 
bounded  by  arcs  of  great  circles.  These  spherical  polygons 
are  named  triangles,  quadrilaterals,  pentagons,  etc.,  accord- 
ing to  the  number  of  sides,  just  as  are  plane  polygons. 

The  nomenclature  is  exactly  the  same  for  the  parts  of  a 
spherical  polygon,  as  for  plane  polygons.  It  is  to  be  borne 
in  mind,  only,  that  the  lines  are  all  great  circle  arcs,  and 
hence  are  usually  expressed  in  circular  units,  not  in  linear 
units.  It  is  to  be  understood,  that  a  great  circle  arc  bears 
exactly  the  same  relation  to  a  spherical  surface  that  a 
straight  line  does  to  a  plane  surface.  In  fact  a  great  circle 
arc  is  really  a  straight  line  bent  to  conform  to  a  spherical 
surface.  In  consequence  a  great  circle  arc  can  be  readily 
drawn  on  a  sphere  with  a  straight  ruler,  if  it  is  flexible 
enough  to  conform  to  the  surface.  Observe  that  a  great 
circle  arc  is  the  shortest  distance  between  two  points  on  a 
sphere,  and  compare  this  fact  with  the  similar  character  of 
a  straight  line. 

Proposition  LVII. 

A  spherical  angle  is  measured  by  the  arc  of  the  great 
circle  described  with  its  vertex  as  pole  and  included  between 
its  sides  (or  sides  produced). 


Solid  Geometry.  199 

Statement:  The  spherical  angle  BAG  is  measured  by  arc 
BC,  described  from  pole  A. 

Analysis:  In  determining  the  value  of  the 
angle  A,  by  definition,  the  angle  between 
the  tangents  EA  and  FA,  drawn  to  the  arcs 
BA  and  CA,  respectively,  decides.  By  com- 
paring this  with  a  central  angle,  which  BC 
subtends,  the  arc  BC  is  introduced. 

Proof:  The  spherical  angle  A  =  Z  EAF 
between  the  tangents  EA  and  FA  (by  definition);  also  the 
diameter  AOD,  drawn  from  A,  is  perpendicular  to  the 
plane  of  the  circle  BCH,  of  which  BC  is  an  arc  (Why?). 
Hence  it  is  perpendicular  to  OB  and  OC,  radii  of  BCH. 
But  EA  and  FA  are  both  perpendicular  to  AO  (Why?). 
Hence  EA  and  AF  are  ||  respectively  to  BO  and  OC, 
since  they  lie  in  the  same  planes.  /.  Z  EAF  = 
Z  BOC,  and  Z  BOC  is  measured  by  arc  BC.  Hence 
Z  EAF  =  arc  BC. 

Cor.  A  spherical  angle  has  the  same  measure  as  the 
dihedral  angle  formed  by  the  planes  of  its  sides. 

Proposition  LVIII. 

One  side  of  a  spherical  triangle  is  less  than  the  sum  of 
the  other  two. 

Statement:  Let  bac  be  a  spherical  triangle  on  the  surface 
of  the  sphere  O,  and  ac  the  largest  side  say;  then  ac  <  ab 
+  be. 

Analysis:  The  sides  of  a  spherical  triangle  being  arcs  of 
circles  are  measured  by  the  angle  at  the  centre  of  the  sphere 
that  they  subtend. 

It  is  evident  that  if  the  three  vertices  of  a  spherical  tri- 
angle be  connected  to  the  centre  of  the  sphere  by  radii, 


200  Solid  Geometry, 

these  radii  will  form  the  edges  of  a  trihedral  angle  at  O, 
whose  faces  will  be  the  planes  of  the 
sides. 

Recalling  the  proposition  referring 
to  the  relative  size  of  one  face  angle 
'of  a  trihedral  to  the  sum  of  the  other 
two,  complete  the  proof. 

Exercise:  Prove  also  by  the  use  of 
the   polyhedral   angle   at  the  centre 
Fig.6io  Of  the  sphere,  that   the   sum   of   the 

sides  of  a  spherical  polygon  is  less  than  four  right  angles. 
Definition:  If  with  the  vertices  of  a  spherical  tri- 
angle as  poles  intersecting  arcs  of  great  circles  be  de- 
scribed, they  will  form  another  spherical  triangle.  These 
two  triangles  are  called  polar  triangles  with  reference  to 
each  other. 

If  with  the  vertices  of  a  spherical  triangle,  as  poles  com- 
plete great  circles  are  described,  they  will,  by  their  inter- 
section, divide  the  surface  of  the  sphere  into  eight  spherical 
triangles. 

If  the  planes  of  these  circles  are  mutually  perpendicular 
the  surface  is  divided  into  eight  trirectangular  *  triangles. 
Each  of  these  triangles  is  divided  into  90  spherical  degrees 
as  they  are  called;  that  is,  the  entire  surface  of  the  sphere 
is  divided  inio  720  spherical  degrees.  These  degrees  are 
used  as  units  of  spherical  surface.  A  spherical  degree 
must  be  carefully  distinguished  from  ordinary  degrees,  as 
spherical  degrees  are  small  unit  surfaces. 

Definition:  The  spherical  excess  is  the  name  given  to  the 
number  of  degrees  by  which  the  sum  of  the  angles  of  a 
spherical  triangle  exceeds  180°.     Evidently  the  trirectang- 
ular triangles  might  be  divided  into  any  number  of  parts, 
*  See  remark  under  Proposition  LX. 


Solid  Geometry. 


201 


as  90  is  purely  arbitrary,  but  it  has  been  found  most  con- 
venient to  make  the  number  of  parts  90. 

Proposition  LIX. 

In  two  polar  triangles  each  angle 
of  one  is  the  supplement  of  the 
opposite  side  in  the  other. 

Statement:  If  ABC  is  a  spheri- 
cal triangle  and  A'B'C'  its  polar, 
then  denoting  the  sides  opposite 
the  angles,  by  the  small  letters, 
corresponding  to  the  large  letters  representing  the  angle. 

A  +  a!  =  180    B   +  V  =  180,  etc. 
Also   A'  +  a    =  1 80     B'  +  b  =  180,  etc. 

Analysis:  As  usual  when  two  figures  are  to  be  compared 
they  are  joined  together  in  some  simple,  direct  way.  The 
plan  that  is  readily  suggested  here,  is  to  produce  the  sides 
of  the  inner  triangle  ABC  to  meet  the  side,  say  B'C',  of 
A'B'C'. 

Proof:  Let  AB  and  AC  be  produced  (along  a  great  circle 
of  course)  to  meet  B'C'  at  E  and  D,  respectively.  Then 
B'D  and  C'E  are  quadrants,  hence, 

B'D  +  C'E  =  180°,     but     B'D  -  B'E  +  ED 
and  C'E  =  C'D  +  ED. 

/.  B'E  +  ED  +  C'D  +  ED  =  180 

(B'E  +  ED  +  C'D  -  B'C'). 

/.  B'C'  +  ED  =  180. 

But  ED  measures  the  angle  A  (by  Prop.  LVII)  and 
B'C'  =  a',  hence  a'  +  A  =  180.  In  the  same  way  by 
extending  the  other  sides  the  other  relations  expressed 
above  may  be  shown. 


2O2  Solid  Geometry. 

Proposition  LX. 

The  sum  of  the  angles  of  a  spherical  triangle  is  greater 
than  1 80°  and  less  than  540°. 

Statement:  To  prove  that  ZA  +  ZB  +  ZCis>  180° 
and  <  540°  in  triangle  ABC. 

Analysis:  Recalling  that  the  sum  of  the  sides  of  any 
spherical  triangle  is  less  than  360°; 
and  also  remembering  the  relation 
between  the  sides  and  angles  of 
polar  triangles,  the  employment  of 
the  triangle  polar  to  ABC  is  imme- 
diately suggested. 

Proof:  Construct  the  triangle 
A'B'C',  polar  to  ABC.  Represent 
as  usual  the  sides  by  the  small 
letters  corresponding  to  the  large  angle-letters  opposite. 
Then  (by  Prop.  LIX) 

A  +  a'  =  180° 
B  +  V  =  180° 
C  +  c'  =  180° 

Add;  (A  +  B+  C)  +  (a'  +  V  +  c')  -  540°; 

transposing      (A  +  B  +  C)  =  540°  -  (a'  +  b'  +  c'). 

But  a'  +  b'  +  c'  <  360°  (by  Exercise  under  Prop. 
LVIII). 

.'.  A  +  B  +  C  =  540°  minus  a  quantity  less  than  360°. 
/.  A  +  B  +  C  >  180°  (since  540°  =  360°  +  180°). 
a'  +  bf  +  c'  cannot  be  o,  or  there  would  be  no  triangle. 
/.  A  +  B  +  C  <  540°. 

Remark:  Unlike  plane  triangles  spherical  triangles  may 
have  more  than  one  right  angle.  Draw  one  with  three 
right  angles.  If  a  spherical  triangle  has  two  right  angles, 
it  is  called  bi-rectangular;  if  it  has  three  it  is  called  tri- 
rectangular. 


Solid  Geometry.  203 

Definitions. 

The  propositions  already  proved  for  plane  triangles  can 
be  verified  in  the  case  of  spherical  triangles  by  identical 
processes,  in  particular  by  superposition.  For  instance, 
two  triangles  on  the  same  sphere  or  equal  spheres  are 
equal  if  two  sides  and  the  included  angle  in  one  are  equal 
to  two  sides  and  the  included  angle  in  the  other,  etc. 

A  zone  is  a  portion  of  the  surface  of  a  sphere  included 
between  two  parallel  planes.  The  circumferences  of  the 
sections  are  called  bases  and  the  perpendicular  between 
the  planes  is  the  altitude. 

If  a  sphere  is  cut  by  a  single  plane,  either  portion  of 
the  surface,  so  divided,  is  called  a  zone  of  one  base. 

A  lime  is  a  portion  of  the  surface  of  a  sphere,  bounded  by 
the  semi-circumferences  of  two  great  circles.  The  angle 
between  these  two  semi-circumferences  is  called  the  angle 
of  the  lune. 

Proposition  LXI. 

The  area  oj  the  surface  generated  by  a  straight  line  revolv- 
ing about  an  axis  in  its  plane  is  equal  to  the  product  of  the 
projection  oj  the  line  on  the  axis  by  the  circwnference,  whose 
radius  is  a  perpendicular  erected  at  the  middle  point  oj  the 
line  and  terminated  by  the  axis. 

Statement:  Let  BF  be  the  line  revolving  about  the  axis 
ocy,  KH  its  projection  on  xy,  and  CE  the  perpendicular  at 
its  mid-point,  terminating  in  xy,  then 

surface  BF  =  KH  X  27rCE. 

Analysis:  There  are  evidently  three  cases:  when  the 
line  BF  is  ||  to  xy]  when  it  is  not  parallel,  but  does  not, 
intersect  xy;  and  when  it  does  meet  xy. 

In  the  first  case  the  surface  generated  is  clearly  that  of 


2O4 


Solid  Geometry. 


a  right  cylinder,  and  the  perpendicular  to  the  middle  of 

the  line  is  the  same  as  the  radius  of  the  bases.     Fig.  64  (M). 

In  the  other  cases  it  is  necessary  to  determine  the  kind 

of  surface  generated;  and  having  decided  that,  apply  the 


(M) 
C 


JJ     K  D     E     H    (K) 

Fig.  64- 


E  H 


corresponding  rule  for  area;  then  by  similar  triangles  re- 
duce the  expression  obtained  for  area  to  the  expression 
required  by  the  proposition. 

Proof:  Case  II.     Fig.  64  (N). 

The  surface  is  that  of  a  frustrum  of  a  cone.  Hence 
surface  BF  =  BF  X  2;rCD,  where  CD  is  the  radius  of 
the  mid-section. 

Draw  BG  ||  to  xy  and  J_  to  FGH,  the  radius  of  larger 
base  of  frustrum. 

Then  BFG  and  DCE  are  similar  right  triangles.     (Why  ?) 
/.  BF  :  CE  :  :  BG  :  CD. 

But    BG  =  KH  (Why?).     .',  BF  :  CE  :  :  KH  :  CD. 
Multiplying  extremes  and  means : 

BF  X  CD  =  CE  X  KH.  (i). 

Surface  BF  =  BF  X  27rCD  may  be  expressed  thus: 

Surface  BF  =  BF  X  CD  X  27T.  (2), 

Substituting  value  of  BF  X  CD  from  (i)  in  (2), 

Surface    BF  =  CE  X  KH  X  271  =  KH  X  27rCE. 

Proof:  Case  III. 

Here  B  coincides  with  K  and  the  surface  is  that  of  a  cone. 

Hence  surface  BF  =  i  (BF  X  2;rFH)  =  BF  X  TrFH(a) 

The  triangles  CDE  and  BFH  are  similar.  (Why?) 


Solid  Geometry.  205 

BF:CE:  :BH  :  CD; 

hence  BF  X  CD  =  CE  X  BH, 

but  CD  =  i  FH  or  FH  =  2CD; 

hence  BF    X  i  FH  =  CE  X  BH 

or  BF    X  FH  =  2CE  X  BH     (b). 

Substituting  this  value  of  BF  X  FH  from  (b)  in  (a), 
Surface    BF  =  BH  X  2?rCE    or  KH  X  2?rCE,   since   B 

and  K  are  same  point. 

Proposition  LXII. 

The  area  of  the  surface  of  a  sphere  is  equal  to  the  product 
of  its  diameter  by  the  circumference  of  a  great  circle. 

Statement:  If  O  be  a  sphere  of  diameter  mt,  then  surface 

Of  O  =   mt  X    27TW0. 

Analysis:  Since  any  curve  may  be  regarded  as  made  up 
of  an  infinite  number  of  infinitely  small  straight  lines, 
and  hence  a  circle  may  be  considered  as  a  polygon  of 
infinite  number  of  sides,  taking  "a  section  through  the  centre 
of  the  sphere  O,  we  may  inscribe  in  the  section  (which  is 
a  circle)  a  regular  polygon  of  any  number  of  sides.  If 
then  half  of  this  polygon  is  revolved  about  the  diameter  as 
an  axis,  each  side  will  generate  a  surface,  which  can  be 
determined  by  Prop.  LXI. 

Proof:    Inscribe  in  the  semi-circle  mabt  the  half  of  a 


Fig.  65- 

regular  hexagon  (a  hexagon  is  the  easiest  regular  polygon 


•2o6  Solid  Geometry. 

to  inscribe)  and  from  the  vertices  a  and  b  drop  perpen- 
diculars to  determine  the  projections  of  the  sides  ma,  ab 
and  bt  on  the  diameter  mt;  also 

Draw  co  perpendicular  to  ma  at  its  middle  point  c, 
Draw  do  perpendicuar  to  ab     at  its  middle  point  d, 
Draw  oe  perpendicular  to  bt    at  its  middle  point  e. 
Then       mj  is  the  projection  of  ma  on   mt. 
Then        jg  is  the  projection  of  ab  on  mt, 
and  gt  is  the  projection  of  bt    on  w/. 

If  the  figure  is  revolved  about  mt  as  an  axis,  the  semi- 
circle will  describe  the  sphere  O,  and  each  side  of  the  semi- 
polygon  will  describe  an  area,  under  the  conditions  of 
Prop.  LXI. 

.'.  area  described  by  ma  =  mj  X  2nco 
area  described  by  ab  =  jg  X  27ido 
area  described  by  bt    =  gt  X  2noe 
add;        area  described  by  (ma  +  ab  +  bt)  = 

(mj  +  jg  +  gt)  X  27zco  [since  co  =  do  —  oe\  that  is,  area 
described  by  mabt  =  mt  X  2nco  [since  mf  +  fg  +  g/  =  mt\. 

If  now  the  sides  of  the  polygon  be  indefinitely  increased 
in  number,  the  sum  of  their  projections  will  continue  to 
be  mt,  since  the  extremities  m  and  /  may  remain  as  they 
are  and  co  will  become  longer  and  longer  until  it  equals 
the  radius  in  length,  when  the  semi-perimeter  of  the  polygon 
becomes  identical  with  the  semi-circumference. 

But  the  above  equation  is  not  confined  to  any  particular 
polygon,  for,  although  the  original  polygon  is  a  hexagon, 
the  fact  that  it  has  six  sides  has  not  been  considered  in  the 
conditions  that  formed  the  equation.  Hence  it  is  true 
for  any  inscribed  polygon,  even  one  which  has  an  infinite 
number  of  sides. 


Solid  Geometry.  207 

.*.  area  described  by  semi-polygon  of  infinite  number 
of  sides  =  mt  X  27rR, 

or    area  of  sphere  =  mt  X  2^-R  =  2R  X  2?rR  =  4;rR2, 
since   a  polygon  of  an  infinite  number  of  sides  is  a  circle, 
and  a  semi-circle  describes  a  spherical  surface,  if  revolved 
about  its  diameter. 

Cor.  I.  Surfaces  of  spheres  are  to  each  other  as  the 
squares  of  their  radii. 

Cor.  II.  The  area  of  a  zone  equals  the  product  of  its 
altitude  by  the  circumference  of  a  great  circle.  For  (Fig.  65) 
the  area  described  by  arc  ab  =  zone  ab  =  gf  X  2;rR,  etc. 

Again,  zone  described  by  arc 

ma  =  mf  X  2;rR  =  mf  X  mt  X  n. 
But          mf  X  mt  =  ma"  in  right  triangle  mat 

(by  Plane  Geometry), 
.'.  zone  described  by  ma  =  Ttma  . 

That  is,  the  area  of  a  zone  of  one  base  equals  the  area  of 
a  circle  whose  radius  is  the  chord  of  the  generating  arc. 

Cor.  Ill-  Zones  on  the  same  sphere  or  equal  spheres  are 
to  each  other  as  their  altitudes. 

Proposition  LXIII. 

The  area  of  a  lune,  in  spherical  degrees,  equals  twice  its 
angle. 

Statement:  The  area  of  the  lune  ABCD  on  the  sphere 
O  equals  2A,  in  spherical  degrees. 

Analysis:  It  is  evidently  necessary  to  compare  the  sur- 
face of  the  lune  with  the  entire  spherical  surface,  and  the 
form  of  the  lune  suggests  that  it  can  be  best  compared  by 
its  angle,  or  what  amounts  to  the  same,  by  the  arc  which 
measures  its  angle. 

Proof:  With  A  as  a  pole  describe  the  great  circle  BCE, 


208 


Solid  Geometry. 


intersecting  the  side  of  the  lune  in  B  and  C.     Then  the 
angle  A  is  measured  by  the  arc  BC.  (Why?) 

If  arc  BC  is  commensurable  with  the 
circle  BCE,  say  BC  contains  a  measure,  m 
times,  and  the  whole  circle  BCE,  n  times, 
then  BC  : BCE  ::m:n(i).  Through  these 
division  points  and  A,  draw  great  circles; 
they  will  divide  the  whole  surface  into  n  unit- 
lunes  and  the  lune  ABCD  into  m  unit-lunes. 
/.  If  L  represent  the  area  of  the  lune  and  S  the  spheri- 
cal surface,  L  :  S  :  :  m  :  n  (2)  .'.'  from  (i)  and  (2)  L  :  S 
:  :  BC  :  BCE. 


Fig.  66. 


But 


and 


that  is, 


BC  ==  Z  A  (Prop.  LVIE), 
BCE  =  360° 

S  =  720  spherical  degrees. 
/.     L  :  720  :  :  A  :  360,* 

r2O  A 


L  = 


=  2A  spherical  degrees. 


If  BC  and  BCE  are  incommensurable,  the  proposition 
can  be  proved  as  usual,  by  reducing  the  unit. 

Proposition  LXIV. 

The  area  of  a  spherical  triangle  is  equal  to  its  spherical 
excess,  expressed  in  spherical  degrees. 

Statement:  If  ABC  is  a  spherical  triangle, 
then  area  ABC  =  A  +  B  +  C  -  180  =  E, 
in  spherical  degrees. 

Analysis:  Since  we  know  the  area  of 
lunes,  it  would  be  helpful,  evidently,  to  ex- 
press the  triangle  in  lunes.  To  accom- 

*  Since  A  and  360  are  both  expressed  in  ordinary  degrees,  their 
ratio,  A  :  360,  is  an  absolute  number,  like  4,  6,  etc. 


Fig.  67. 


Solid  Geometry.  209 

plish  this  it  is  necessary  to  complete  the  great  circles,  of 
which  the  sides  of  the  triangles  are  arcs. 

Prooj:  Complete  the  great  circles  as  indicated.  They 
will  intersect  in  the  vertices  A,  B,  and  C,  and  also  in  the 
points  A',  B/  and  C.' 

Then  BCA  +  BCA'  =  lune  ABA'C 
BCA  +  BAG'  =  lune  CBC'A 
BCA  +  CAB'  =  lune  CBAB' 

add;         2BCA  +  BCA  +  BCA'  +  BAG'  +  CAB' 
=  lunes  ABA'C  +  CBC'A  +  CBAB'. 
But  BCA+BCA'+BAC'-{-CAB'=  the  hemisphere 

(since  BCA'  =  B'C'A)* 

and  lunes  ABA'C  +  CBC'A  +  CBAB'  =  2A+2C  +  2B 
Hence,  since  the  hemispheres  =  360  spherical  degrees, 
BCA  +  180  =  A  +  B  +  C  spherical  degrees, 

or         BCA  =  A  +  B  +  C-i8o  =  E  spherical  degrees. 
Proposition  LXV. 

The  volume  0}  a  sphere  equals  one-third  the  radius  multi- 
plied by  the  surface. 

Statement:  Let  X  be  a  sphere  of  radius  R  and  surface 
S,  then  the  volume  V  =  J  RS. 

Analysis:  The  proposition  suggests  the  volume  of  a 
pyramid,  and  regarding  the  surface  of  the  sphere  as  made 
up  of  an  infinite  number  of  plane  faces,  and  connecting  the 
intersection  points  of  these  planes  with  the  centre  of  the 
sphere,  the  analogy  is  complete. 

*  B'C'A  is  called  the  symmetrical  triangle  to  BCA',  formed  by 
drawing  diameters  through  B,  C,  and  A',  and  joining  the  other 
extremities  of  these  diameters  by  arcs  of  great  circles.  Clearly  the 
sides  of  these  triangles  are  equal  in  length,  but  arranged  in  reverse 
order,  hence  they  are  equivalent. 


210  Solid  Geometry. 

Proof:  Circumscribe  a  polygon  of  any  convenient  num- 
ber of  faces  about  the  sphere. 

Join  the  vertices  of  this  polyhedron  with  the  centre  of 
the  sphere,  and  then  pass  planes  through  these  centre 
lines  and  the  edges  of  the  polyhedron.  The  polyhedron  is 
thus  divided  into  pyramids  with  their  apices  at  the  centre; 
and  with  the  faces  of  the  polyhedron  as  bases.  Since  these 
bases  are  tangent  to  the  sphere  the  radius  of  the  sphere 
will  be  the  altitude  of  each  pyramid.  (Why?) 

The  volume  of  each  pyramid  equals  J  its  base,  by  the 
radius,  hence  the  sum  of  all  the  pyramids  will  equal  J  the 
radius  times  the  surface  of  the  polyhedron  (which  is  the 
sum  of  all  the  bases).  If  the  number  of  these  faces  is 
indefinitely  increased,  the  surface  of  the  polyhedron  will 
become  eventually  the  surface  of  the  sphere,  but  the  volume 
is  always  equal  to  J  the  radius  times  the  surface,  hence, 
when  it  coincides  with  the  sphere,  V  =  J  RS. 

Cor.  I.  The  volume  of  the  sphere  may  be  stated  thus: 
Let  R  =  radius;  S,  the  surface;  and  V,  the  volume. 

By  Prop.  LXV,  V  -  J  RS, 
but  S  =  47rR2  (by  Prop.  LXII), 

/.    V  =    |  TlR3  -   J  7TD3, 

since  R  =  J  D,  hence  R3  =  J  D3. 

Cor.  II.  The  volumes  of  two  spheres  are  to  each  other 
as  the  cubes  of  their  radii,  or  as  the  cubes  of  their  diameters. 

Cor.  III.  The  volume  of  a  spherical  sector  is  equal  to 
one-third  the  product  of  the  radius  and  the-  area  of  the 
zone  forming  the  base  of  the  sector. 

EXERCISE    V. 

i.  The  radius  of  a  sphere  is  12  in.  Find  the  area  of 
the  section  made  by  a  plane  bisecting  the  radius  perpen- 
dicularly. 


Solid  Geometry.  211 

2.  What  is  the  locus  of  the  centres  of  spheres  having  a 
given  radius  and  tangent  to  a  given  sphere  ? 

3.  Find  the  locus  of  the  centres  of  spheres  having  a 
given  radius  and  tangent  to  two  given  spheres. 

4.  If  two  straight  lines  are  tangent  to  a  sphere  at  the 
same  point,  their  plane  is  tangent  to  the  sphere  at  that 
point. 

5.  To  draw  a  sphere  tangent  to  four  intersecting  planes. 

6.  Show  how  to  pass  a  sphere  through  four  points  in 
space  not  in  the  same  plane. 

7.  What  is  the  radius  of  a  sphere  whose  surface  is  25,434 
square  metres  ? 

8.  Two  leaden  spheres  of  20  in.  and  40  in.  diameter, 
respectively,  are  to  be  melted  and  a  single  sphere  formed. 
What  is  the  diameter  of  this  sphere  ? 

9.  If  specific  gravity  equals  weight  divided  by  volume, 
what  is  the  diameter  of  an  iron  sphere  whose  weight  is 
100  Ibs.     The  specific  gravity  of  iron  being  7.2. 

10.  How  far  can  an  observer  on  the  ocean  see  if  he  is 
elevated  150  ft.  above  the  surface,  taking  the  earth's  radius 
to  be  3,960  miles? 

11.  How  many  cubic  feet  of  masonry  will  it  take  to  build 
a  dome  in  the  form  of  a  half-sphere  whose  diameter  is  21  ft. 
and  thickness  15  in.? 

12.  Through  a  given  line  to  pass  a  plane  tangent  to  a 
given  sphere. 

13.  To  inscribe  a  sphere  in  a  given  tetrahedron. 

14.  What  is  the  radius  of  a  sphere  inscribed  in  a  regular 
tetrahedron,  each  of  whose  edges  is  3  \/6  ? 

15.  On  a  sphere  whose  radius  is  10  in.,  a  small  circle 
of  radius  5  \/3  is  described.      How  far  is  its  plane  from 
the  centre  ? 


212  Solid  Geometry. 

16.  Show  how  to  circumscribe  a  circle  about  a  spherical 
triangle. 

17.  The  diameter  of  a  sphere  is  17  in.     From  any  point 
on  the  surface  as  a  pole,  with  the  points  of  the  compass 
spread  8  in.,  a  circle  is  described  on  the  spherical  surface. 
Find  its  area. 

1 8.  Find  the  radius  of  the  section  of  a  sphere  13  in.  in 
diameter  made  5  in.  from  the  centre. 

19.  Construct  a  spherical  surface  with  a  given  radius 
that  passes  through  two  given  points  and  touches  a  given 
sphere. 

20.  The  angles  of  a  spherical  triangle  are  125°,  110°  and 
146°.     What  is  its  area  in  square  feet  if  the  radius  of  the 
sphere  is  7  ft. 

21.  What  is  the  area  of  the  circle  of  intersection  of  two 
spheres  of  8  in.  and  6  in.  radius,  respectively,  if  their  centres 
are  5  in.  apart?  - 

22.  The  radii  of  two  parallel  sections  of  a  sphere  are 
6  in.  and  4  in.,  respectively,  and  they  are  2  in.  apart.   What 
is  the  radius  of  the  sphere  ? 

23.  How  high  must  a  man  be  raised  above  the  earth's 
surface  that  he  may  see  J  of  it  ? 

24.  What  is  the  radius  of  a  sphere  whose  volume  is 
equal  numerically  to  the  circumference  of  a  great  circle  ? 

25.  What  is  the  surface  in  square  feet  of  a  lune  whose 
angle  is  36°  on  a  sphere  whose  radius  is  14  ft.  ? 

26.  Find  the  area  of  the  torrid  zone  on  the  earth,  if 
its  altitude  is  3,200  miles,  the  earth's  diameter  being  7,920 
miles. 


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